Exercise 7.5 (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths
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NCERT Class 12 Maths Solutions: Chapter 7 Integrals | Free PDF Download
Ex 7.5 Question 1.
$\frac{x}{(x+1)(x+2)}$
Answer.
$\frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}$.
$
\begin{aligned}
& \Rightarrow x=\mathrm{A}(x+2)+\mathrm{B}(x+1) \\
& \Rightarrow x=\mathrm{A} x+2 \mathrm{~A}+\mathrm{B} x+\mathrm{B}
\end{aligned}
$
Comparing coefficients of $x$ on both sides $\mathrm{A}+\mathrm{B}=1$
Comparing constants $2 \mathrm{~A}+\mathrm{B}=0$
Solving eq. (ii) and (iii), we get $A=-1$ and $B=2$
Putting these values of A and B in eq. (i),
$
\begin{aligned}
& \frac{x}{(x+1)(x+2)}=\frac{-1}{x+1}+\frac{2}{x+2} \\
& \therefore \int \frac{x}{(x+1)(x+2)} d x=-\int \frac{1}{x+1} d x+2 \int \frac{1}{x+2} d x \\
& =-\log |x+1|+2 \log |x+2|+c \\
& =\log |x+2|^2-\log |x+1|+c
\end{aligned}
$
$
=\log \frac{(x+2)^2}{|x+1|}+c
$
Ex 7.5 Question 2.
$\frac{1}{x^2-9}$
$
\begin{aligned}
& \text { Answer. } \int \frac{1}{x^2-9} d x \\
& =\int \frac{1}{x^2-3^2} d x \\
& =\frac{1}{2 \times 3} \log \left|\frac{x-3}{x+3}\right|+c \\
& {\left[\because \int \frac{1}{x^2-a^2} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\right]} \\
& =\frac{1}{6} \log \left|\frac{x-3}{x+3}\right|+c
\end{aligned}
$
Ex 7.5 Question 3.
$\frac{3 x-1}{(x-1)(x-2)(x-3)}$
Answer.
$\frac{3 x-1}{(x-1)(x-2)(x-3)}$
$
\begin{aligned}
& =\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x-3} \ldots \ldots . .(\mathrm{i}) \\
& \Rightarrow 3 x-1=\mathrm{A}(x-2)(x-3)+\mathrm{B}(x-1)(x-3)+\mathrm{C}(x-1)(x-2) \\
& \Rightarrow 3 x-1=\mathrm{A}\left(x^2-5 x+6\right)+\mathrm{B}\left(x^2-4 x+3\right)+\mathrm{C}\left(x^2-3 x+2\right) \\
& \Rightarrow 3 x-1=\mathrm{A} x^2-5 \mathrm{~A} x+6 \mathrm{~A}+\mathrm{B} x^2-4 \mathrm{~B} x+3 \mathrm{~B}+\mathrm{C} x^2-3 \mathrm{C} x+2 \mathrm{C}
\end{aligned}
$
Comparing coefficients of $x^2: \mathrm{A}+\mathrm{B}+\mathrm{C}=0$
$\text { Comparing coefficients of } x:-5 \mathrm{~A}-4 \mathrm{~B}-3 \mathrm{C}=3$
$
\Rightarrow 5 \mathrm{~A}+4 \mathrm{~B}+3 \mathrm{C}=-3
$
Comparing constants: $6 \mathrm{~A}+3 \mathrm{~B}+2 \mathrm{C}=-1$ $\qquad$
On solving eq. (i), (ii) and (iii), we get $\mathrm{A}=1, \mathrm{~B}=-5, \mathrm{C}=4$
Putting the values of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in eq. (i),
$
\begin{aligned}
& \frac{3 x-1}{(x-1)(x-2)(x-3)} \\
& =\frac{1}{x-1}+\frac{-5}{x-2}+\frac{4}{x-3} \\
& \Rightarrow \int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x \\
& =\int \frac{1}{x-1} d x-5 \int \frac{1}{x-2} d x+4 \int \frac{1}{x-3} d x \\
& \Rightarrow \int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x \\
& =\log |x-1|-5 \log |x-2|+4 \log |x-3|+c
\end{aligned}
$
Ex 7.5 Question 4.
$\frac{x}{(x-1)(x-2)(x-3)}$
$
\begin{aligned}
& \text { Answer. } \frac{x}{(x-1)(x-2)(x-3)} \\
& =\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x-3} \ldots \ldots . .(\mathrm{i}) \\
& \Rightarrow x=\mathrm{A}(x-2)(x-3)+\mathrm{B}(x-1)(x-3)+\mathrm{C}(x-1)(x-2)
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow x=\mathrm{A}\left(x^2-5 x+6\right)+\mathrm{B}\left(x^2-4 x+3\right)+\mathrm{C}\left(x^2-3 x+2\right) \\
& \Rightarrow x=\mathrm{A} x^2-5 \mathrm{~A} x+6 \mathrm{~A}+\mathrm{B} x^2-4 \mathrm{~B} x+3 \mathrm{~B}+\mathrm{C} x^2-3 \mathrm{C} x+2 \mathrm{C}
\end{aligned}
$
Comparing coefficients of $x^2: A+B+C=0$
Comparing coefficients of $x:-5 \mathrm{~A}-4 \mathrm{~B}-3 \mathrm{C}=1$
$
\Rightarrow 5 \mathrm{~A}+4 \mathrm{~B}+3 \mathrm{C}=-1
$
Comparing constants: $6 \mathrm{~A}+3 \mathrm{~B}+2 \mathrm{C}=0$ $\qquad$ (iv)
On solving eq. (i), (ii) and (iii), we get $A=\frac{1}{2}, B=-2, C=\frac{3}{2}$
Putting the values of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in eq. (i),
$
\frac{x}{(x-1)(x-2)(x-3)}
$
$
\begin{aligned}
& =\frac{1 / 2}{x-1}+\frac{-2}{x-2}+\frac{3 / 2}{x-3} \\
& \Rightarrow \int \frac{x}{(x-1)(x-2)(x-3)} d x \\
& =\frac{1}{2} \int \frac{1}{x-1} d x-2 \int \frac{1}{x-2} d x+\frac{3}{2} \int \frac{1}{x-3} d x \\
& \Rightarrow \int \frac{x}{(x-1)(x-2)(x-3)} d x \\
& =\frac{1}{2} \log |x-1|-2 \log |x-2|+\frac{3}{2} \log |x-3|+c
\end{aligned}
$
Ex 7.5 Question 5.
$\frac{2 x}{x^2+3 x+2}$
$\begin{aligned}
& \text { Answer. } \frac{2 x}{x^2+3 x+2} \\
& =\frac{2 x}{x^2+2 x+x+2} \\
& =\frac{2 x}{x(x+2)+1(x+2)} \\
& =\frac{2 x}{(x+1)(x+2)} \\
& =\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2} \cdots(\mathrm{i}) \\
& \Rightarrow 2 x=\mathrm{A}(x+2)+\mathrm{B}(x+1) \\
& \Rightarrow 2 x=\mathrm{A} x+2 \mathrm{~A}+\mathrm{B} x+\mathrm{B}
\end{aligned}$
Comparing coefficients of $x$ on both sides $\mathrm{A}+\mathrm{B}=2$ $\qquad$
Comparing constants $2 \mathrm{~A}+\mathrm{B}=0$ $\qquad$
Solving eq. (ii) and (iii), we get $\mathrm{A}=-2$ and $\mathrm{B}=4$
Putting these values of $A$ and $B$ in eq. (i),
$
\begin{aligned}
& \frac{2 x}{(x+1)(x+2)}=\frac{-2}{x+1}+\frac{4}{x+2} \\
& \therefore \int \frac{2 x}{(x+1)(x+2)} d x=-2 \int \frac{1}{x+1} d x+4 \int \frac{1}{x+2} d x \\
& =-2 \log |x+1|+4 \log |x+2|+c \\
& =4 \log |x+2|-2 \log |x+1|+c
\end{aligned}
$
Ex 7.5 Question 6.
$\frac{1-x^2}{x(1-2 x)}$
Answer.
$\frac{1-x^2}{x(1-2 x)}$
$
\begin{aligned}
& =\frac{1-x^2}{x-2 x^2} \\
& =\frac{-x^2+1}{-2 x^2+x} \\
& =\frac{1}{2}+\frac{\left(-\frac{x}{2}+1\right)}{x(1-2 x)}
\end{aligned}
$
$=\frac{1}{2}+\frac{\left(-\frac{x}{2}+1\right)}{x(1-2 x)}$
(Dividing numerator by denominator)
$
\therefore \int \frac{1-x^2}{x(1-2 x)} d x=\int\left\{\frac{1}{2}+\frac{\left(-\frac{x}{2}+1\right)}{x(1-2 x)}\right\} d x
$
$
=\frac{1}{2} \int 1 d x+\int \frac{-\frac{x}{2}+1}{x(1-2 x)} d x
$
Now $\int \frac{-\frac{x}{2}+1}{x(1-2 x)} d x$
$
=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{1-2 x}
$
$
\Rightarrow-\frac{x}{2}+1=\mathrm{A}(1-2 x)+\mathrm{B} x
$
$
\Rightarrow-\frac{x}{2}+1=\mathrm{A}-2 \mathrm{~A} x+\mathrm{B} x
$
Comparing coefficients of $x$ on both sides $-2 \mathrm{~A}+\mathrm{B}=\frac{-1}{2} \ldots .$. (iii)
Comparing constants $\mathrm{A}=1$ .(iv) $\qquad$
Solving eq. (ii) and (iii), we get $A=1$ and $B=\frac{3}{2}$
Putting these values of $A$ and $B$ in eq. (ii),
$
\begin{aligned}
& \frac{-\frac{x}{2}+1}{x(1-2 x)}=\frac{1}{x}+\frac{3 / 2}{1-2 x} \\
& \therefore \int \frac{-\frac{x}{2}+1}{x(1-2 x)} d x=\int \frac{1}{x} d x+\frac{3}{2} \int \frac{1}{1-2 x} d x \\
& =\log |x|+\frac{3}{2} \log \frac{|1-2 x|}{-2 \rightarrow \text { Coeff. of } x}+c \\
& =\log |x|-\frac{3}{4} \log |1-2 x|+c \\
&
\end{aligned}
$
Putting this value in eq. (i),
$
\begin{aligned}
& \int \frac{1-x^2}{x(1-2 x)} d x \\
& =\frac{1}{2} x+\log |x|-\frac{3}{4} \log |1-2 x|+c
\end{aligned}
$
Ex 7.5 Question 7.
$\frac{x}{\left(x^2+1\right)(x-1)}$
Answer.
$\frac{x}{\left(x^2+1\right)(x-1)}=\frac{A x+B}{x^2+1}+\frac{C}{x-1}$
$
\begin{aligned}
& \Rightarrow x=(\mathrm{A} x+\mathrm{B})(x-1)+\mathrm{C}\left(x^2+1\right) \\
& \Rightarrow x=\mathrm{A} x^2-\mathrm{A} x+\mathrm{B} x-\mathrm{B}+\mathrm{C} x^2+\mathrm{C}
\end{aligned}
$
Comparing coefficients of $x^2, \mathrm{~A}+\mathrm{C}=0$
Comparing coefficients of $x,-\mathrm{A}+\mathrm{B}=1$
Comparing constant terms, $-\mathrm{B}+\mathrm{C}=0$ .(iv)
Solving eq. (ii), (iii) and (iv), we get $A=\frac{-1}{2}, B=\frac{1}{2}$ and $C=\frac{1}{2}$
Putting the values of A, B and C in eq. (i), $\frac{x}{\left(x^2+1\right)(x-1)}=\frac{\frac{-1}{2} x+\frac{1}{2}}{x^2+1}+\frac{\frac{1}{2}}{x-1}$
$
\begin{aligned}
& \Rightarrow \frac{x}{\left(x^2+1\right)(x-1)}=\frac{-1}{2} \cdot \frac{x}{x^2+1}+\frac{1}{2} \cdot \frac{1}{x^2+1}+\frac{1}{2} \cdot \frac{1}{x-1} \\
& =\frac{-1}{4} \cdot \frac{2 x}{x^2+1}+\frac{1}{2} \cdot \frac{1}{x^2+1}+\frac{1}{2} \cdot \frac{1}{x-1} \\
& \Rightarrow \int \frac{x}{\left(x^2+1\right)(x-1)} d x=\frac{-1}{4} \int \frac{2 x}{x^2+1} d x+\frac{1}{2} \int \frac{1}{x^2+1} d x+\frac{1}{2} \int \frac{1}{x-1} d x \\
& \Rightarrow \int \frac{x}{\left(x^2+1\right)(x-1)} d x=\frac{-1}{4} \log \left|x^2+1\right|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+c
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \int \frac{x}{\left(x^2+1\right)(x-1)} d x=\frac{-1}{4} \log \left(x^2+1\right)+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+c \\
& \Rightarrow \int \frac{x}{\left(x^2+1\right)(x-1)} d x=\frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^2+1\right)+\frac{1}{2} \tan ^{-1} x+c
\end{aligned}
$
Ex 7.5 Question 8.
$\frac{x}{(x-1)^2(x+2)}$
$
\begin{aligned}
& \text { Answer. } \frac{x}{(x-1)^2(x+2)} \\
& =\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{(x-1)^2}+\frac{\mathrm{C}}{x+2} \ldots \ldots . \text { (i) } \\
& \Rightarrow x=\mathrm{A}(x-1)(x+2)+\mathrm{B}(x+2)+\mathrm{C}(x-1)^2 \\
& \Rightarrow x=\mathrm{A}\left(x^2+x-2\right)+\mathrm{B}(x+2)+\mathrm{C}\left(x^2-2 x+1\right) \\
& \Rightarrow x=\mathrm{A} x^2+\mathrm{A} x-2 \mathrm{~A}+\mathrm{B} x+2 \mathrm{~B}+\mathrm{C} x^2-2 \mathrm{C} x+\mathrm{C}
\end{aligned}
$
Comparing coefficients of $x^2: \mathrm{A}+\mathrm{C}=0$. ..(ii)
Comparing coefficients of $x: \mathrm{A}+\mathrm{B}-2 \mathrm{C}=1$ $\qquad$
Comparing constants: $-2 \mathrm{~A}+2 \mathrm{~B}+\mathrm{C}=0$ $\qquad$ .(iv)
On solving eq. (i), (ii) and (iii), we get
$
\mathrm{A}=\frac{2}{9}, \mathrm{~B}=\frac{1}{3}, \mathrm{C}=\frac{-2}{9}
$
Putting the values of A, B and C in eq. (i), $\frac{x}{(x-1)^2(x+2)}$
$\begin{aligned}
& =\frac{\frac{2}{9}}{x-1}+\frac{\frac{1}{3}}{(x-1)^2}+\frac{\frac{-2}{9}}{x+2} \\
& \Rightarrow \int \frac{x}{(x-1)^2(x+2)} d x \\
& =\frac{2}{9} \int \frac{1}{x-1} d x+\frac{1}{3} \int \frac{1}{(x-1)^2} d x-\frac{2}{9} \int \frac{1}{x+2} d x \\
& =\frac{2}{9} \log |x-1|+\frac{1}{3} \int(x-1)^{-2} d x-\frac{2}{9} \log |x+2|+c \\
& =\frac{2}{9} \log |x-1|+\frac{1}{3} \frac{(x-1)^{-1}}{(-1)(1)}-\frac{2}{9} \log |x+2|+c \\
& =\frac{2}{9}(\log |x-1|-\log |x+2|)-\frac{1}{3(x-1)}+c \\
& =\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+c
\end{aligned}$
Ex 7.5 Question 9.
$\frac{3 x+5}{x^3-x^2-x+1}$
Answer.
$\frac{3 x+5}{x^3-x^2-x+1}$
$
\begin{aligned}
& =\frac{3 x+5}{x^2(x-1)-1(x-1)} \\
& =\frac{3 x+5}{(x-1)\left(x^2-1\right)}
\end{aligned}
$
$
\begin{aligned}
& =\frac{3 x+5}{(x-1)(x-1)(x+1)} \\
& =\frac{3 x+5}{(x-1)^2(x+1)} \\
& =\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{(x-1)^2}+\frac{\mathrm{C}}{x+1} \ldots \ldots . .(\mathrm{i}) \\
& \Rightarrow 3 x+5=\mathrm{A}(x-1)(x+1)+\mathrm{B}(x+1)+\mathrm{C}(x-1)^2 \\
& \Rightarrow 3 x+5=\mathrm{A}\left(x^2-1\right)+\mathrm{B}(x+1)+\mathrm{C}\left(x^2-2 x+1\right) \\
& \Rightarrow 3 x+5=\mathrm{A} x^2-\mathrm{A}+\mathrm{B} x+\mathrm{B}+\mathrm{C} x^2-2 \mathrm{C} x+\mathrm{C}
\end{aligned}
$
Comparing coefficients of $x^2: \mathrm{A}+\mathrm{C}=0$
Comparing coefficients of $x: \mathrm{B}-2 \mathrm{C}=3$
Comparing constants: $-2 \mathrm{~A}+\mathrm{B}+\mathrm{C}=5$ $\qquad$
On solving eq. (i), (ii) and (iii), we get $A=\frac{-1}{2}, B=4, C=\frac{1}{2}$
Putting the values of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in eq. (i),
$\begin{aligned}
& \frac{3 x+5}{x^3-x^2-x+1} \\
& =\frac{\frac{-1}{2}}{x-1}+\frac{4}{(x-1)^2}+\frac{\frac{1}{2}}{x+1} \\
& =\frac{-1}{2} \int \frac{1}{x-1} d x+4 \int(x-1)^{-2} d x+\frac{1}{2} \int \frac{1}{x+1} d x
\end{aligned}$
$
\begin{aligned}
& =\frac{-1}{2} \log |x-1|+4 \frac{(x-1)^{-1}}{(-1)(1)}+\frac{1}{2} \log |x+1|+c \\
& =\frac{1}{2}(\log |x+1|-\log |x-1|)-\frac{4}{(x-1)}+c \\
& =\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{(x-1)}+c
\end{aligned}
$
Ex 7.5 Question 10.
$\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}$
Answer.
$\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}$
$
\begin{aligned}
& =\frac{2 x-3}{(x-1)(x+1)(2 x+3)} \\
& =\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+1}+\frac{\mathrm{C}}{2 x+3} \ldots \ldots . \text { (i) } \\
& \Rightarrow 2 x-3=\mathrm{A}(x+1)(2 x+3)+\mathrm{B}(x-1)(2 x+3)+\mathrm{C}(x-1)(x+1) \\
& \Rightarrow 2 x-3=\mathrm{A}\left(2 x^2+5 x+3\right)+\mathrm{B}\left(2 x^2+x-3\right)+\mathrm{C}\left(x^2-1\right) \\
& \Rightarrow 2 x-3=2 \mathrm{~A} x^2+5 \mathrm{~A} x+3 \mathrm{~A}+2 \mathrm{~B} x^2+\mathrm{B} x-3 \mathrm{~B}+\mathrm{C} x^2-\mathrm{C}
\end{aligned}
$
Comparing coefficients of $x^2: 2 \mathrm{~A}+2 \mathrm{~B}+\mathrm{C}=0$
Comparing coefficients of $x: 5 \mathrm{~A}+\mathrm{B}=2$
Comparing constants: $3 \mathrm{~A}-3 \mathrm{~B}-\mathrm{C}=-3$ $\qquad$
On solving eq. (i), (ii) and (iii), we get $\mathrm{A}=\frac{-1}{10}, \mathrm{~B}=\frac{5}{2}, \mathrm{C}=\frac{-24}{5}$
Putting the values of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in eq. (i),
$\begin{aligned}
& \frac{2 x-3}{\left(x^2-1\right)(2 x+3)} \\
& =\frac{\frac{-1}{10}}{x-1}+\frac{\frac{5}{2}}{x+1}+\frac{\frac{-24}{2}}{2 x+3} \\
& \Rightarrow \int \frac{2 x-3}{\left(x^2-1\right)(2 x+3)} d x=\frac{-1}{10} \int \frac{1}{x-1} d x+\frac{5}{2} \int \frac{1}{x+1} d x-\frac{24}{5} \int \frac{1}{2 x+3} d x \\
& =\frac{-1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{24}{5} \frac{\log |2 x+3|}{2 \rightarrow \operatorname{Coeff} . \text { of } x}+c \\
& =\frac{-1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+c \\
& =\frac{5}{2} \log |x+1|-\frac{1}{10} \log |x-1|-\frac{12}{5} \log |2 x+3|+c
\end{aligned}$
Ex 7.5 Question 11.
$\frac{5 x}{(x+1)\left(x^2-4\right)}$
Answer.
$\frac{5 x}{(x+1)\left(x^2-4\right)}$
$
\begin{aligned}
& =\frac{5 x}{(x+1)(x+2)(x-2)} \\
& =\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C}}{x-2} \ldots \ldots . .(\mathrm{i}) \\
& \Rightarrow 5 x=\mathrm{A}(x+2)(x-2)+\mathrm{B}(x+1)(x-2)+\mathrm{C}(x+1)(x+2) \\
& \Rightarrow x=\mathrm{A}\left(x^2-4\right)+\mathrm{B}\left(x^2-x-2\right)+\mathrm{C}\left(x^2+3 x+2\right) \\
& \Rightarrow x=\mathrm{A} x^2-4 \mathrm{~A}+\mathrm{B} x^2-\mathrm{B} x-2 \mathrm{~B}+\mathrm{C} x^2+3 \mathrm{C} x+2 \mathrm{C}
\end{aligned}
$
Comparing coefficients of $x^2: A++B+C=0$ $\qquad$
Comparing coefficients of $x:-B+3 C=5$ $\qquad$
Comparing constants: $-4 \mathrm{~A}-2 \mathrm{~B}+2 \mathrm{C}=0$ $\qquad$ (iv)
On solving eq. (i), (ii) and (iii), we get $\mathrm{A}=\frac{5}{3}, \mathrm{~B}=\frac{-5}{2}, \mathrm{C}=\frac{5}{6}$
Putting the values of $A, B$ and $C$ in eq. (i),
$
\begin{aligned}
& \frac{5 x}{(x+1)\left(x^2-4\right)} \\
& =\frac{\frac{5}{3}}{x+1}+\frac{\frac{-5}{2}}{(x+2)}+\frac{\frac{5}{6}}{x-2} \\
& =\frac{5}{3} \int \frac{1}{x+1} d x-\frac{5}{2} \int \frac{1}{(x+2)} d x+\frac{5}{6} \int \frac{1}{x-2} d x \\
& =\frac{5}{3} \log |x+1|-\frac{5}{2} \log |x+2|+\frac{5}{6} \log |x-2|+c
\end{aligned}
$
Ex 7.5 Question 12.
$\frac{x^3+x+1}{x^2-1}$
Answer.
$\frac{x^3+x+1}{x^2-1}$
$=x+\frac{2 x+1}{x^2-1}$
[On dividing numerator by denominator]
Let $\frac{2 x+1}{x^2-1}$
$
\begin{aligned}
& =\frac{2 x+1}{(x+1)(x-1)} \\
& =\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x-1} \ldots \ldots . . \text { (ii) } \\
& \Rightarrow 2 x+1=\mathrm{A}(x-1)+\mathrm{B}(x+1) \\
& \Rightarrow 2 x+1=\mathrm{A} x-\mathrm{A}+\mathrm{B} x+\mathrm{B}
\end{aligned}
$
Comparing coefficients of $x: \mathrm{A}+\mathrm{B}=2$ $\qquad$
Comparing constants: $-\mathrm{A}+\mathrm{B}=1$ $\qquad$ .(iv)
On solving eq. (iii) and (iv), we get $\mathrm{A}=\frac{1}{2}, \mathrm{~B}=\frac{3}{2}$
Putting the values of A, B and C in eq. (ii), $\frac{2 x+1}{x^2-1}$
$
=\frac{1 / 2}{x+1}+\frac{3 / 2}{x-1}
$
Putting this value in eq. (i),
$
\begin{aligned}
& \frac{x^3+x+1}{x^2-1} \\
& =x+\frac{1 / 2}{x+1}+\frac{3 / 2}{x-1} \\
& \Rightarrow \int \frac{x^3+x+1}{x^2-1} d x \\
& =\int x d x+\frac{1}{2} \int \frac{1}{x+1} d x+\frac{3}{2} \int \frac{1}{x-1} d x
\end{aligned}
$
$=\frac{x^2}{2}+\frac{1}{2} \log |x+1|+\frac{3}{2} \log |x-1|+c$
Ex 7.5 Question 13.
$\frac{2}{(1-x)\left(1+x^2\right)}$
Answer.
$\frac{2}{(1-x)\left(1+x^2\right)}$
$=\frac{\mathrm{A}}{1-x}+\frac{\mathrm{B} x+\mathrm{C}}{1+x^2}$
$\Rightarrow 2=\mathrm{A}\left(1+x^2\right)+(\mathrm{B} x+\mathrm{C})(1-x)$
$\Rightarrow 2=\mathrm{A}+\mathrm{A} x^2+\mathrm{B} x-\mathrm{B} x^2+\mathrm{C}-\mathrm{C} x$
Comparing the coefficients of $x^2 \mathrm{~A}-\mathrm{B}=0$
Comparing the coefficients of $x \mathrm{~B}-\mathrm{C}=0$
Comparing constants $\mathrm{A}+\mathrm{C}=2$ .(iv)
On solving eq. (ii), (iii) and (iv), we get $A=1, B=1, C=1$
Putting these values of $A, B$ and $C$ in eq. (i),
$
\begin{aligned}
& \frac{2}{(1-x)\left(1+x^2\right)}=\frac{1}{1-x}+\frac{x+1}{1+x^2} \\
& =\frac{1}{1-x}+\frac{x}{1+x^2}+\frac{1}{1+x^2}=\frac{1}{1-x}+\frac{1}{2} \cdot \frac{2 x}{1+x^2}+\frac{1}{1+x^2} \\
& =\int \frac{1}{1-x} d x+\frac{1}{2} \int \frac{2 x}{1+x^2} d x+\int \frac{1}{1+x^2} d x
\end{aligned}
$
$\begin{aligned}
& =\frac{\log |1-x|}{-1 \rightarrow \text { Coeff. of } x}+\frac{1}{2} \log \left|1+x^2\right|+\tan ^{-1} x+c \\
& =-\log |1-x|+\frac{1}{2} \log \left(1+x^2\right)+\tan ^{-1} x+c
\end{aligned}$
Ex 7.5 Question 14.
$\frac{3 x-1}{(x+2)^2}$
Answer.
Let $\mathrm{I}=\int \frac{3 x-1}{(x+2)^2} d x$
Putting $x+2=t$
$
\begin{aligned}
& \Rightarrow x=t-2 \\
& \Rightarrow \frac{d x}{d t}=1 \\
& \Rightarrow d x=d t
\end{aligned}
$
Putting this value in eq. (i),
$
\begin{aligned}
& \mathrm{I}=\int \frac{3(t-2)-1}{(t)^2} d t \\
& =\int \frac{3 t-6-1}{t^2} d t \\
& =\int \frac{3 t-7}{t^2} d t \\
& =\int\left(\frac{3 t}{t^2}-\frac{7}{t^2}\right) d t \\
& =\int\left(\frac{3}{t}-\frac{7}{t^2}\right) d t
\end{aligned}
$
$\begin{aligned}
& =3 \int \frac{1}{t} d t-7 \int t^{-2} d t \\
& =3 \log |t|-7 \frac{t^{-1}}{-1}+c \\
& =3 \log |t|+\frac{7}{t}+c \\
& =3 \log |x+2|+\frac{7}{x+2}+c
\end{aligned}$
Ex 7.5 Question 15.
$\frac{1}{x^4-1}$
Answer.
$\frac{1}{x^4-1}$
$
=\frac{1}{\left(x^2-1\right)\left(x^2+1\right)}
$
Putting $x^2=y, \frac{1}{x^4-1}$
$=\frac{1}{(y-1)(y+1)}=\frac{\mathrm{A}}{y-1}+\frac{\mathrm{B}}{y+1}$
$\Rightarrow 1=\mathrm{A}(y+1)+\mathrm{B}(y-1)$
$\Rightarrow 1=\mathrm{A} y+\mathrm{A}+\mathrm{B} y-\mathrm{B}$
Comparing the coefficients of $y \mathrm{~A}+\mathrm{B}=0$ $\qquad$
Comparing constants $\mathrm{A}-\mathrm{B}=1$ $\qquad$
On solving the eq. (ii) and (iii), we get $\mathrm{A}=\frac{1}{2}, \mathrm{~B}=\frac{-1}{2}$
Putting the values of A, B and $y$ in eq. (i),
$\begin{aligned}
& \frac{1}{x^4-1}=\frac{\frac{1}{2}}{x^2-1}+\frac{\frac{-1}{2}}{x^2+1} \\
& \Rightarrow \int \frac{1}{x^4-1} d x=\frac{1}{2} \int \frac{1}{x^2-1} d x-\frac{1}{2} \int \frac{1}{x^2+1} d x \\
& =\frac{1}{2} \cdot \frac{1}{2 \cdot 1} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+c \\
& =\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+c
\end{aligned}$
Ex 7.5 Question 16.
$\frac{1}{x\left(x^n+1\right)}$
Answer.
Let I $=\int \frac{1}{x\left(x^n+1\right)} d x$
Multiplying both numerator and denominator by $n x^{n-1}$,
$
\begin{aligned}
& {\left[\because \frac{d}{d x}\left(x^n+1\right)=n x^{n-1}\right]} \\
& \mathrm{I}=\int \frac{n x^{n-1}}{n x^{n-1} x\left(x^n+1\right)} d x \\
& =\frac{1}{n} \int \frac{n x^{n-1}}{x^n\left(x^n+1\right)} d x \ldots \ldots . . \text { (i) }
\end{aligned}
$
Putting $x^n=t$
$
\begin{aligned}
& \Rightarrow n x^{n-1}=\frac{d t}{d x} \\
& \Rightarrow n x^{n-1} d x=d t
\end{aligned}
$
$\begin{aligned}
& =\frac{1}{n} \int \frac{1}{t(t+1)} d t \\
& =\frac{1}{n} \int \frac{t+1-t}{t(t+1)} d t \\
& =\frac{1}{n} \int \frac{t+1}{t(t+1)}-\frac{t}{t(t+1)} d t \\
& =\frac{1}{n}\left[\int \frac{1}{t} d t+\int \frac{1}{t+1} d t\right] \\
& =\frac{1}{n}[\log |t|-\log |t+1|]+c \\
& =\frac{1}{n}\left[\log \left|x^n\right|-\log \left|x^n+1\right|\right]+c \\
& =\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|+c
\end{aligned}$
Ex 7.5 Question 17.
$\frac{\cos x}{(1-\sin x)(2-\sin x)}$
Answer.
Let $\mathrm{I}=\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x \ldots \ldots$. (i)
Putting $\sin x=t$
$
\Rightarrow \cos x=\frac{d t}{d x}
$
$\begin{aligned}
& \Rightarrow \cos x d x=d t \\
& \therefore \text { From eq. (i), } \mathrm{I}=\int \frac{1}{(1-t)(2-t)} d t \\
& =\int \frac{(2-t)-(1-t)}{(1-t)(2-t)} d t \\
& =\int\left(\frac{(2-t)}{(1-t)(2-t)}-\frac{(1-t)}{(1-t)(2-t)}\right) d t \\
& =\int\left(\frac{1}{(1-t)}-\frac{1}{(2-t)}\right) d t \\
& =\int \frac{1}{1-t} d t-\int \frac{1}{2-t} d t \\
& =\frac{\log |1-t|}{-1-\operatorname{Coeff} \cdot \text { of } t}-\frac{\log |2-t|}{-1}+c \\
& =-\log |1-t|+\log |2-t|+c \\
& =\log \left|\frac{2-t}{1-t}\right|+c \\
& =\log \left|\frac{2-\sin x}{1-\sin x}\right|+c
\end{aligned}$
Ex 7.5 Question 18.
$\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}$
Answer.
$\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} \ldots \ldots$. (i)
Putting $x^2=y$,
$
\begin{aligned}
& \frac{(y+1)(y+2)}{(y+3)(y+4)} \\
& =\frac{y^2+3 y+2}{y^2+7 y+12}
\end{aligned}
$
Dividing numerator by denominator,
$
\begin{aligned}
& \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} \\
& =1+\frac{(-4 y-10)}{(y+3)(y+4)} \ldots . \text { (iii) } \\
& \text { Let } \frac{(-4 y-10)}{(y+3)(y+4)}=\frac{\mathrm{A}}{y+3}+\frac{\mathrm{B}}{y+4} \ldots . \\
& \Rightarrow-4 y-10=\mathrm{A}(y+4)+\mathrm{B}(y+3) \\
& \Rightarrow-4 y-10=\mathrm{A} y+4 \mathrm{~A}+\mathrm{B} y+3 \mathrm{~B}
\end{aligned}
$
Let $\frac{(-4 y-10)}{(y+3)(y+4)}=\frac{A}{y+3}+\frac{B}{y+4}$
Comparing coefficients of $y \mathrm{~A}+\mathrm{B}=-4$ $\qquad$
Comparing constants $4 \mathrm{~A}+3 \mathrm{~B}=-10$ $\qquad$ .(vi)
On solving eq. (v) and (vi), we get $A=2, B=-6$
Putting the values of A, B and $y$ in eq. (iii),
$
\begin{aligned}
& \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} \\
& =1+\frac{2}{x^2+3}-\frac{6}{x^2+4}
\end{aligned}
$
$\begin{aligned}
& \Rightarrow \int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x \\
& =\int\left(1+\frac{2}{x^2+3}-\frac{6}{x^2+4}\right) d x \\
& =\int 1 d x+2 \int \frac{1}{x^2+(\sqrt{3})^2} d x-6 \int \frac{1}{x^2+2^2} d x \\
& =x+2 \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-6 \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+c \\
& =x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+c
\end{aligned}$
Ex 7.5 Question 19.
$\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}$
Answer.
Let I $=\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x$
Putting $x^2=t \Rightarrow 2 x d x=d t$
$\therefore$ From eq. (i),
$
\begin{aligned}
& \mathrm{I}=\int \frac{d t}{(t+1)(t+3)} \\
& =\frac{1}{2} \int \frac{2}{(t+1)(t+3)} d t \\
& =\frac{1}{2} \int \frac{(t+3)-(t+1)}{(t+1)(t+3)} d t
\end{aligned}
$
$\begin{aligned}
& =\frac{1}{2} \int\left(\frac{(t+3)}{(t+1)(t+3)}-\frac{(t+1)}{(t+1)(t+3)}\right) d t \\
& =\frac{1}{2} \int\left(\frac{1}{(t+1)}-\frac{1}{(t+3)}\right) d t \\
& =\frac{1}{2}[\log |t+1|-\log |t+3|]+c \\
& =\frac{1}{2}\left[\log \left|\frac{t+1}{t+3}\right|\right]+c \\
& =\frac{1}{2} \log \left|\frac{x^2+1}{x^2+3}\right|+c \\
& =\frac{1}{2} \log \left(\frac{x^2+1}{x^2+3}\right)+c
\end{aligned}$
Ex 7.5 Question 20.
$\frac{1}{x\left(x^4-1\right)}$
Answer.
Let $I=\int \frac{1}{x\left(x^4-1\right)} d x$
$
\begin{aligned}
& =\int \frac{4 x^3}{4 x^3 x\left(x^4-1\right)} d x \\
& =\frac{1}{4} \int \frac{4 x^3}{x^4\left(x^4-1\right)} d x \ldots \text { (i) } \\
& {\left[\because \frac{d}{d x}\left(x^4-1\right)=4 x^3\right]}
\end{aligned}
$
Putting $x^4=t$
$
\begin{aligned}
& \Rightarrow 4 x^3=\frac{d t}{d x} \\
& \Rightarrow 4 x^3 d x=d t
\end{aligned}
$
Putting this value in eq. (i),
$
\begin{aligned}
\mathrm{I} & =\frac{1}{4} \int \frac{d t}{t(t-1)} \\
& =\frac{1}{4} \int \frac{t-(t-1)}{t(t-1)} d t \\
\mathrm{I} & =\frac{1}{4} \int\left(\frac{t}{t(t-1)}-\frac{(t-1)}{t(t-1)}\right) d t \\
& =\frac{1}{4} \int\left(\frac{1}{(t-1)}-\frac{1}{t}\right) d t \\
& =\frac{1}{4}\left[\int \frac{1}{t-1} d t-\int \frac{1}{t} d t\right] \\
& =\frac{1}{4}[\log |t-1|-\log |t|]+c \\
& =\frac{1}{4} \log \left|\frac{t-1}{t}\right|+c \\
& =\frac{1}{4} \log \left|\frac{x^4-1}{x^4}\right|+c
\end{aligned}
$
Ex 7.5 Question 21
$\frac{1}{\left(e^x-1\right)}$
Answer.
Let $\mathrm{I}=\int \frac{1}{e^x-1} d x$
Putting $e^x=t$
$
\begin{aligned}
& \Rightarrow e^x=\frac{d t}{d x} \\
& \Rightarrow e^x d x=d t \\
& \Rightarrow d x=\frac{d t}{e^x}
\end{aligned}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\int \frac{1}{t-1} \frac{d t}{e^x} \\
& =\int \frac{1}{t-1} \frac{d t}{t} \\
& =\int \frac{1}{t(t-1)} d t \\
& =\int \frac{t-(t-1)}{t(t-1)} d t \\
& =\int\left(\frac{t}{t(t-1)}-\frac{(t-1)}{t(t-1)}\right) d t
\end{aligned}
$
$\begin{aligned}
& =\int\left(\frac{1}{(t-1)}-\frac{1}{t}\right) d t \\
& =\int \frac{1}{t-1} d t-\int \frac{1}{t} d t \\
& =\log |t-1|-\log |t|+c \\
& =\log \left|\frac{t-1}{t}\right|+c
\end{aligned}$
$=\log \left|\frac{e^x-1}{e^x}\right|+c$
Ex 7.5 Question 22.
$\int \frac{x d x}{(x-1)(x-2)}$ equals:
(A) $\log \left|\frac{(x-1)^2}{x-2}\right|+C$
(B) $\log \left|\frac{(x-2)^2}{x-1}\right|+C$
(C) $\left.\log \left(\frac{x-1}{x-2}\right)^2 \right\rvert\,+C$
(D) $\log |(x-1)(x-2)|+C$
Answer.
Let $\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$
$\Rightarrow x=\mathrm{A}(x-2)+\mathrm{B}(x-1)$
$\Rightarrow x=\mathrm{A} x-2 \mathrm{~A}+\mathrm{B} x-\mathrm{B}$
Comparing coefficients of $x \mathrm{~A}+\mathrm{B}=1$
Comparing constants $-2 \mathrm{~A}-\mathrm{B}=0$
On solving eq. (ii) and (iii), we get $\mathrm{A}=-1, \mathrm{~B}=2$
Putting these values of A and B in eq. (i),
$
\frac{x}{(x-1)(x-2)}=\frac{-1}{x-1}+\frac{2}{x-2}
$
$
\begin{aligned}
& \Rightarrow \int \frac{x}{(x-1)(x-2)} d x=-\int \frac{1}{x-1} d x+2 \int \frac{1}{x-2} d x \\
& =-\log |x-1|+2 \log |x-2|+c \\
& =\log \left|(x-2)^2\right|-\log |x-1|+c \\
& =\log \left|\frac{(x-2)^2}{x-1}\right|+c
\end{aligned}
$
Therefore, option (B) is correct.
Ex 7.5 Question 23.
$\int \frac{d x}{x\left(x^2+1\right)}$ equals:
(A) $\log |x|-\frac{1}{2} \log \left(x^2+1\right)+C$
(B) $\log |x|+\frac{1}{2} \log \left(x^2+1\right)+C$
(C) $-\log |x|-\frac{1}{2} \log \left(x^2+1\right)+$ C
(D) $\frac{1}{2} \log |x|+\log \left(x^2+1\right)+\mathrm{C}$
Answer.
Let $\mathrm{I}=\int \frac{1}{x\left(x^2+1\right)} d x$
$
\begin{aligned}
& =\int \frac{2 x}{2 x x\left(x^2+1\right)} d x \\
& =\int \frac{2 x}{2 x^2\left(x^2+1\right)} d x .
\end{aligned}
$
$
\left[\because \frac{d}{d x}\left(x^2+1\right)=2 x\right]
$
Putting $x^2=t$
$
\begin{aligned}
& \Rightarrow 2 x=\frac{d t}{d x} \\
& \Rightarrow 2 x d x=d t
\end{aligned}
$
Putting this value in eq. (i),
$
\begin{aligned}
\mathrm{I} & =\int \frac{d t}{2 t(t+1)} \\
& =\frac{1}{2} \int \frac{(t+1)-t}{t(t+1)} d t \\
\mathrm{I} & =\frac{1}{2} \int\left(\frac{t+1}{t(t+1)}-\frac{t}{t(t+1)}\right) d t \\
& =\frac{1}{2} \int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t \\
& =\frac{1}{2}\left[\int \frac{1}{t} d t-\int \frac{1}{t+1} d t\right] \\
& =\frac{1}{2}[\log |t|-\log |t+1|]+c \\
& =\frac{1}{2}\left[2 \log \left|x^2\right|-\log \left(x^2+1\right)\right]+c
\end{aligned}
$
$
=\log |x|-\frac{1}{2} \log \left(x^2+1\right)+c
$
Therefore, option (A) is correct.
