Exercise 7.4 (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 7.4 Question 1.
$\frac{3 x^2}{x^6+1}$
Answer.
Let $\mathrm{I}=\int \frac{3 x^2}{x^6+1} d x$
$
=\int \frac{3 x^2}{\left(x^3\right)^2+1} d x
$
Putting $x^3=t$
$
\begin{aligned}
& \Rightarrow 3 x^2=\frac{d t}{d x} \\
& \Rightarrow 3 x^2 d x=d t \\
& \therefore \text { From eq. (i), } \\
& \mathrm{I}=\int \frac{d t}{t^2+1} \\
& =\frac{1}{1} \tan ^{-1} \frac{t}{1}+c \quad \text { using } \int \frac{1}{a^2+x^2} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c \\
& =\tan ^{-1} x^3+c \text { Ans. }
\end{aligned}
$
Ex 7.4 Question 2.
$\frac{1}{\sqrt{1+4 x^2}}$
Answer.
$\int \frac{1}{\sqrt{1+4 x^2}} d x$
$
\begin{aligned}
& =\int \frac{1}{\sqrt{(2 x)^2+1^2}} d x \\
& =\frac{\log \left|(2 x)+\sqrt{(2 x)^2+1^2}\right|}{2 \rightarrow \text { Coeff. of } x}+c \\
& {\left[\because \int \frac{1}{\sqrt{x^2+a^2}} d x=\log \left|x+\sqrt{x^2+a^2}\right|\right]} \\
& =\frac{1}{2} \log \left|2 x+\sqrt{4 x^2+1}\right|+c
\end{aligned}
$
Ex 7.4 Question 3.
$\frac{1}{\sqrt{(2-x)^2+1}}$
$
\begin{aligned}
& \text { Answer. } \int \frac{1}{\sqrt{(2-x)^2+1}} d x \\
& =\frac{\log \mid(2-x)+\sqrt{(2-x)^2+1^2}}{-1 \rightarrow \text { Coeff. of } x}+c \\
& {\left[\because \int \frac{1}{\sqrt{x^2+a^2}} d x=\log \left|x+\sqrt{x^2+a^2}\right|\right]} \\
& =-\log \left|2-x+\sqrt{4+x^2-4 x+1}\right|+c \\
& ==\log \left|\frac{1}{2-x+\sqrt{x^2-4 x+5}}\right|+c
\end{aligned}
$
Ex 7.4 Question 4.
$\frac{1}{\sqrt{9-25 x^2}}$
$
\begin{aligned}
& \text { Answer. } \int \frac{1}{\sqrt{9-25 x^2}} d x \\
& =\int \frac{1}{\sqrt{(3)^2-(5 x)^2}} d x \\
& =\frac{\sin ^{-1} \frac{5 x}{3}}{5 \rightarrow \operatorname{Coeff} . \text { of } x}+c \\
& {\left[\because \int \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1} \frac{x}{a}\right]} \\
& =\frac{1}{5} \sin ^{-1}\left(\frac{5 x}{3}\right)+c
\end{aligned}
$
Ex 7.4 Question 5.
$\frac{3 x}{1+2 x^4}$
Answer.
Let $\mathrm{I}=\int \frac{3 x}{1+2 x^4} d x$
$=\frac{3}{2} \int \frac{2 x}{1+2\left(x^2\right)^2} d x$
Putting $x^2=t$
$\Rightarrow 2 x=\frac{d t}{d x}$
$\Rightarrow 2 x d x=d t$
$\therefore$ From eq. (i),
$\begin{aligned}
& \mathrm{I}=\frac{3}{2} \int \frac{d t}{1+2 t^2} \\
& =\frac{3}{2} \int \frac{1}{(\sqrt{2} t)^2+1^2} d t \\
& =\frac{3}{2} \frac{\frac{1}{1} \tan ^{-1} \frac{\sqrt{2} t}{1}}{\sqrt{2} \rightarrow \text { Coeff. of } t}+c \\
& ==\frac{3}{2 \sqrt{2}} \tan ^{-1} \sqrt{2} t+c \\
& ==\frac{3}{2 \sqrt{2}} \tan ^{-1} \sqrt{2} x^2+c \text { Ans. } \\
&
\end{aligned}$
Ex 7.4 Question 6.
$\frac{x^2}{1-x^6}$
Answer.
Let $\mathrm{I}=\int \frac{x^2}{1-x^6} d x$
$
\begin{aligned}
& =\int \frac{x^2}{1-\left(x^3\right)^2} d x \\
& =\frac{1}{3} \int \frac{3 x^2}{1-\left(x^3\right)^2} d x .
\end{aligned}
$
Putting $x^3=t$
$
\begin{aligned}
& \Rightarrow 3 x^2=\frac{d t}{d x} \\
& \Rightarrow 3 x^2 d x=d t
\end{aligned}
$
$\therefore$ From eq. (i),
$\begin{aligned}
& \mathrm{I}=\frac{1}{3} \int \frac{d t}{1-t^2} \\
& ==\frac{1}{3} \int \frac{1}{(1)^2-(t)^2} d t \\
& =\frac{1}{3} \cdot \frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|+c \\
& {\left[\because \int \frac{1}{a^2-x^2} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\right]} \\
& =\frac{1}{6} \log \left|\frac{1+x^3}{1-x^3}\right|+c
\end{aligned}$
Ex 7.4 Question 7.
$\frac{x-1}{\sqrt{x^2-1}}$
Answer.
Let $\mathrm{I}=\int \frac{x-1}{\sqrt{x^2-1}} d x$
$
\begin{aligned}
& =\int\left(\frac{x}{\sqrt{x^2-1}}-\frac{1}{\sqrt{x^2-1}}\right) d x \\
& =\int \frac{x}{\sqrt{x^2-1}} d x-\int \frac{1}{\sqrt{x^2-1}} d x \\
& =\frac{1}{2} \int \frac{2 x}{\sqrt{x^2-1}} d x-\log \left|x+\sqrt{\left(x^2-(1)^2\right)}\right|
\end{aligned}
$
Let $\mathrm{I}_1=\int \frac{2 x}{\sqrt{x^2-1}} d x$
Putting $x^2-1=t$
$
\Rightarrow 2 x=\frac{d t}{d x}
$
$\begin{aligned}
& \Rightarrow 2 x d x=d t \\
& \therefore \mathrm{I}_1=\int \frac{d t}{\sqrt{t}}=\int t^{\frac{-1}{2}} d t=\frac{t^{1 / 2}}{1 / 2}=2 \sqrt{t} \\
& \Rightarrow \mathrm{I}=\frac{1}{2}\left(2 \sqrt{x^2-1}+c\right)-\log \left|x+\sqrt{x^2-1}\right| \\
& =\sqrt{x^2-1}+\frac{c}{2}-\log \left|x+\sqrt{x^2-1}\right| \\
& \Rightarrow \mathrm{I}=\sqrt{x^2-1}-\log \left|x+\sqrt{x^2-1}\right|+c_1 \text { where } c_1=\frac{c}{2}
\end{aligned}$
Ex 7.4 Question 8.
$\frac{x^2}{\sqrt{x^6+a^6}}$
Answer.
Let $\mathrm{I}=\int \frac{x^2}{\sqrt{x^6+a^6}} d x$
$
=\frac{1}{3} \int \frac{x^2}{\sqrt{\left(x^3\right)^2+a^6}} d x \text {. }
$
Putting $x^3=t$
$\Rightarrow 3 x^2=\frac{d t}{d x}$
$\Rightarrow 3 x^2 d x=d t$
$\therefore$ From eq. (i),
$
\mathrm{I}=\frac{1}{3} \int \frac{d t}{t^2+a^6}
$
$\begin{aligned}
& =\frac{1}{3} \int \frac{1}{t^2+\left(a^3\right)^2} d t \\
& =\frac{1}{3} \log \left|t+\sqrt{t^2+\left(a^3\right)^2}\right|+c \\
& =\frac{1}{3} \log \left|x^3+\sqrt{x^6+a^6}\right|+c
\end{aligned}$
Ex 7.4 Question 9.
$\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}$
Answer.
Let $I=\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}} d x$
Putting $\tan x=t$
$
\begin{aligned}
& \Rightarrow \sec ^2 x=\frac{d t}{d x} \\
& \Rightarrow \sec ^2 x d x=d t \\
& \therefore \text { From eq. (i), }
\end{aligned}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
& \mathrm{I}=\int \frac{d t}{\sqrt{t^2+4}} \\
& =\int \frac{1}{\sqrt{t^2+(2)^2}} d t \\
& =\log \left|t+\sqrt{t^2+(2)^2}\right|+c \\
& =\log \left|\tan x+\sqrt{\tan ^2 x+4}\right|+c
\end{aligned}
$
Ex 7.4 Question 10.
$\text {} \frac{1}{\sqrt{x^2+2 x+2}}$
Answer.
$\text {} \int \frac{1}{\sqrt{x^2+2 x+2}} d x$
$\begin{aligned}
& =\int \frac{1}{\sqrt{x^2+2 x+1+1}} d x \\
& =\int \frac{1}{\sqrt{(x+1)^2+(1)^2}} d x \\
& =\log \left|x+1+\sqrt{(x+1)^2+(1)^2}\right|+c \\
& =\log \left|x+1+\sqrt{x^2+2 x+2}\right|+c
\end{aligned}$
Ex 7.4 Question 11.
$\frac{1}{9 x^2+6 x+5}$
Answer.
$\int \frac{1}{9 x^2+6 x+5} d x$
$
\begin{aligned}
& =\int \frac{1}{9\left(x^2+\frac{6 x}{9}+\frac{5}{9}\right)} d x \\
& =\int \frac{1}{9\left(x^2+\frac{2 x}{3}+\frac{5}{9}\right)} d x \\
& =\int \frac{1}{9\left(x^2+\frac{2 x}{3}+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2+\frac{5}{9}\right)} d x
\end{aligned}
$
[For making completing the squares]
$\begin{aligned}
& =\int \frac{1}{9\left\{\left(x+\frac{1}{3}\right)^2+\frac{4}{9}\right\}} d x \\
& =\int \frac{1}{9\left\{\left(x+\frac{1}{3}\right)^2+\left(\frac{2}{3}\right)^2\right\}} d x \\
& =\frac{1}{9} \int \frac{1}{\left\{\left(x+\frac{1}{3}\right)^2+\left(\frac{2}{3}\right)^2\right\}} d x \\
& =\frac{1}{9} \cdot \frac{1}{\left(\frac{2}{3}\right)} \tan ^{-1} \frac{x+\frac{1}{3}}{\frac{2}{3}}+c \\
& {\left[\because \int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]} \\
& =\frac{1}{9} \cdot \frac{3}{2} \tan ^{-1}\left(\frac{\frac{3 x+1}{3}}{\frac{2}{3}}\right)+c \\
& =\frac{1}{6} \tan ^{-1}\left(\frac{3 x+1}{2}\right)+c \\
&
\end{aligned}$
Ex 7.4 Question 12.
$\frac{1}{\sqrt{7-6 x-x^2}}$
Answer.
$\int \frac{1}{\sqrt{7-6 x-x^2}} d x$
$\begin{aligned}
& =\int \frac{1}{\sqrt{-x^2-6 x+7}} d x \\
& =\int \frac{1}{\sqrt{-\left(x^2+6 x-7\right)}} d x \\
& =\int \frac{1}{\sqrt{-\left(x^2+6 x+9-9-7\right)}} d x \\
& =\int \frac{1}{\sqrt{-\left\{(x+3)^2-16\right\}}} d x \\
& =\int \frac{1}{\sqrt{-(x+3)^2+16}} d x \\
& =\int \frac{1}{\sqrt{(4)^2-(x+3)^2}} d x \\
& =\sin ^{-1}\left(\frac{x+3}{4}\right)+c \\
& {\left[\because \int \frac{1}{a^2-x^2} d x=\sin ^{-1} \frac{x}{a}\right]}
\end{aligned}$
Ex 7.4 Question 13.
$\begin{aligned}
& \text {} \frac{1}{\sqrt{(x-1)(x-2)}} \\
& \text { Answer. } \int \frac{1}{\sqrt{(x-1)(x-2)}} d x \\
& =\int \frac{1}{\sqrt{x^2-2 x-x+2}} d x \\
& =\int \frac{1}{\sqrt{x^2-3 x+2}} d x
\end{aligned}$
$\begin{aligned}
& =\int \frac{1}{\sqrt{x^2-3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+2}} d x \\
& =\int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}} d x \\
& =\log \left|x-\frac{3}{2}+\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right|+c \\
& =\log \left|x-\frac{3}{2}+\sqrt{x^2-3 x+2}\right|+c
\end{aligned}$
Ex 7.4 Question 14.
$\frac{1}{\sqrt{8+3 x-x^2}}$
$
\begin{aligned}
& \text { Answer. } \int \frac{1}{\sqrt{8+3 x-x^2}} d x \\
& =\int \frac{1}{\sqrt{-x^2+3 x+8}} d x \\
& =\int \frac{1}{\sqrt{-\left(x^2-3 x-8\right)}} d x \\
& =\int \frac{1}{\sqrt{-\left\{x^2-3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2-8\right\}}} d x
\end{aligned}
$
$\begin{aligned}
& =\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^2-\frac{41}{4}\right\}}} d x \\
& =\int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}} d x \\
& =\sin ^{-1} \frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}+c \\
& =\sin ^{-1}\left(\frac{2 x-3}{\sqrt{41}}\right)+c
\end{aligned}$
Ex 7.4 Question 15.
$\text {} \frac{1}{\sqrt{(x-a)(x-b)}}$
Answer.
$\begin{aligned}
& \text {} \int \frac{1}{\sqrt{(x-a)(x-b)}} d x \\
& =\int \frac{1}{\sqrt{x^2-b x-a x+a b}} d x \\
& =\int \frac{1}{\sqrt{x^2-x(a+b)+a b}} d x \\
& =\int \frac{1}{\sqrt{x^2-x(a+b)+\left(\frac{a+b}{2}\right)^2-\left(\frac{a+b}{2}\right)^2+a b}} d x
\end{aligned}$
$\begin{aligned}
& =\int \frac{1}{\sqrt{\left.\left\{x-\left(\frac{a+b}{2}\right)\right\}^2-\left\{\left(\frac{a-b}{2}\right)^2\right\}\right]}} d x \\
& =\log \left|x-\left(\frac{a+b}{2}\right)+\sqrt{\left\{\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\right\}}\right|+c \\
& =\log \left|x-\left(\frac{a+b}{2}\right)+\sqrt{x^2-x(a+b)+a b}\right|+c \\
&
\end{aligned}$
Ex 7.4 Question 16.
$\frac{4 x+1}{\sqrt{2 x^2+x-3}}$
Answer.
Let I $=\int \frac{4 x+1}{\sqrt{2 x^2+x-3}} d x$ ...(i)
Putting $2 x^2+x-3=t$
$
\begin{aligned}
& \Rightarrow 4 x+1=\frac{d t}{d x} \\
& \Rightarrow(4 x+1) d x=d t
\end{aligned}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\int \frac{d t}{\sqrt{t}} \\
& =\int t^{\frac{-1}{2}} d t \\
& =\frac{t^{1 / 2}}{1 / 2}+c \\
& =2 \sqrt{t}+c \\
& =2 \sqrt{2 x^2+x-3}+c
\end{aligned}
$
Ex 7.4 Question 17.
$\frac{x+2}{\sqrt{x^2-1}}$
Answer.
Let I $=\int \frac{x+2}{\sqrt{x^2-1}} d x$
$
\begin{aligned}
& =\int\left(\frac{x}{\sqrt{x^2-1}}+\frac{2}{\sqrt{x^2-1}}\right) d x \\
& =\int \frac{x}{\sqrt{x^2-1}} d x+2 \int \frac{1}{\sqrt{x^2-1}} d x \\
& =\int \frac{x}{\sqrt{x^2-1}} d x+2 \log \left|x+\sqrt{x^2-1^2}\right|+c
\end{aligned}
$
$
\begin{aligned}
& \text { Let } \mathrm{I}_1=\int \frac{x}{\sqrt{x^2-1}} d x \\
& =\frac{1}{2} \int \frac{2 x}{\sqrt{x^2-1}} d x
\end{aligned}
$
Putting $x^2-1=t$
$
\begin{aligned}
& \Rightarrow 2 x=\frac{d t}{d x} \\
& \Rightarrow 2 x d x=d t \\
& \therefore \mathrm{I}_1=\frac{1}{2} \int \frac{d t}{\sqrt{t}} \\
& =\frac{1}{2} \int t^{\frac{-1}{2}} d t \\
& =\frac{1}{2} \cdot \frac{t^{1 / 2}}{1 / 2}+c \\
& =\sqrt{t} \\
& =\sqrt{x^2-1}
\end{aligned}
$
Putting this value in eq. (i),
$
\sqrt{x^2-1}+2 \log \left|x+\sqrt{x^2-1^2}\right|+c
$
Ex 7.4 Question 18.
$\frac{5 x-2}{1+2 x+3 x^2}$
Answer.
Let I $=\int \frac{5 x-2}{1+2 x+3 x^2} d x$ ...(i)
Let $n$ umerator $=A \frac{d}{d x}($ deno $\min$ ator $)+B$
$
\begin{aligned}
& \Rightarrow 5 x-2=A \frac{d}{d x}\left(1+2 x+3 x^2\right)+B \\
& \Rightarrow 5 x-2=A(2+6 x)+B \ldots . . \text { (ii) } \\
& \Rightarrow 5 x-2=2 A+6 A x+B
\end{aligned}
$
Comparing coefficients of $x, 6 \mathrm{~A}=5$
$
\Rightarrow A=\frac{5}{6}
$
Comparing constants,
$
2 \mathrm{~A}+\mathrm{B}=-2
$
On solving, we get $\mathrm{A}=\frac{5}{6}, \mathrm{~B}=\frac{-11}{3}$
Putting the values of A and B in eq. (ii),
$
5 x-2=\frac{5}{6}(2+6 x)-\frac{11}{3}
$
Putting this value of $5 x-2$ in eq. (i),
$
\begin{aligned}
& I=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^2} d x \\
& \mathrm{I}=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^2} d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^2} d x \\
& \Rightarrow \mathrm{I}=\frac{5}{6} \mathrm{I}_1-\frac{11}{3} \mathrm{I}_2 \\
& \text { Now } \mathrm{I}_1=\int \frac{2+6 x}{1+2 x+3 x^2} d x \\
&
\end{aligned}
$
Putting $1+2 x+3 x^2=t$
$
\begin{aligned}
& \Rightarrow 2+6 x=\frac{d t}{d x} \\
& \Rightarrow(2+6 x) d x=d t \\
& \therefore \mathrm{I}_1=\int \frac{d t}{t}=\log |t|=\log \left|1+2 x+3 x^2\right|
\end{aligned}
$
$
\begin{aligned}
& \text { Again } \mathrm{I}_2=\int \frac{1}{1+2 x+3 x^2} d x \\
& =\int \frac{1}{3 x^2+2 x+1} d x \\
& =\int \frac{1}{3\left(x^2+\frac{2}{3} x+\frac{1}{3}\right)} d x \\
& =\int \frac{1}{3\left[x^2+\frac{2}{3} x+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2+\frac{1}{3}\right]} d x \\
& =\int \frac{1}{3\left[\left(\left(x+\frac{1}{3}\right)^2+\frac{2}{9}\right)\right]} d x \\
& =\frac{1}{3} \int \frac{1}{\left[\left(\left(x+\frac{1}{3}\right)^2+\frac{2}{9}\right)\right]} d x
\end{aligned}
$
$\frac{1}{3} \int \frac{1}{\left[\left(x+\frac{1}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2\right]} d x$
$
\begin{aligned}
& {\left[\because \int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]} \\
& =\frac{1}{3} \cdot \frac{1}{\frac{\sqrt{2}}{3}} \tan ^{-1} \frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}} \\
& =\frac{1}{3} \cdot \frac{3}{\sqrt{2}} \tan ^{-1} \frac{3 x+1}{\sqrt{2}} \\
& =\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right) \ldots \ldots \ldots .(v)
\end{aligned}
$
Putting values of $I_1$ and $I_2$ in eq. (iii),
$
I=\frac{5}{6} \log \left|1+2 x+3 x^2\right|-\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+c
$
Ex 7.4 Question 19.
$\frac{6 x+7}{\sqrt{(x-5)(x-4)}}$
Answer.
Let $\mathrm{I}=\int \frac{6 x+7}{\sqrt{(x-5)(x-4)}} d x$
$
=\int \frac{6 x+7}{\sqrt{x^2-9 x+20}} d x
$
Let $\quad$ Linear $=\mathrm{A} \frac{d}{d x}$ (Quadratic) + B
$
\Rightarrow 6 x+7=\mathrm{A} \frac{d}{d x}\left(x^2-9 x+20\right)+\mathrm{B}
$
$
\begin{aligned}
& \Rightarrow 6 x+7=A(2 x-9)+B \\
& \Rightarrow 6 x+7=2 A x-9 A+B
\end{aligned}
$
Comparing coefficients of $x, 2 \mathrm{~A}=6 \Rightarrow \mathrm{A}=3$
Comparing constants, $-9 \mathrm{~A}+\mathrm{B}=7$
On solving, we get $A=3, B=34$
Putting the values of A and B in eq. (ii),
$
6 x+7=3(2 x-9)+34
$
Putting this value of $6 x+7$ in eq. (i),
$
\begin{aligned}
& \mathrm{I}=\int \frac{3(2 x-9)+34}{\sqrt{x^2-9 x+20}} d x \\
& \mathrm{I}=3 \int \frac{2 x-9}{\sqrt{x^2-9 x+20}} d x+34 \int \frac{1}{\sqrt{x^2-9 x+20}} d x \\
& \Rightarrow \mathrm{I}=3 \mathrm{I}_1+34 \mathrm{I}_2 \ldots \ldots \ldots . . \text { (iii) }
\end{aligned}
$
Now $I_1=\int \frac{2 x-9}{\sqrt{x^2-9 x+20}} d x$
Putting $x^2-9 x+20=t$
$\begin{aligned}
& \Rightarrow 2 x-9=\frac{d t}{d x} \\
& \Rightarrow(2 x-9) d x=d t \\
& \therefore \mathrm{I}_1=\int \frac{d t}{\sqrt{t}} \\
& =\int t^{\frac{-1}{2}} d t
\end{aligned}$
$\begin{aligned}
&\begin{aligned}
& =\frac{t^{1 / 2}}{1 / 2}=2 \sqrt{t} \\
& =2 \sqrt{x^2-9 x+20}
\end{aligned}\\
&\begin{aligned}
& \text { Again } \mathrm{I}_2=\int \frac{1}{\sqrt{x^2-9 x+20}} d x \\
& =\int \frac{1}{\sqrt{x^2-9 x+\left(\frac{9}{2}\right)^2-\left(\frac{9}{2}\right)^2+20}} d x
\end{aligned}
\end{aligned}$
$
\begin{aligned}
& =\int \frac{1}{\sqrt{\left(x-\frac{9}{2}\right)^2-\frac{1}{4}}} d x \\
& =\int \frac{1}{\sqrt{\left(x+\frac{1}{3}\right)^2-\left(\frac{1}{2}\right)^2}} d x \\
& =\log \left|x-\frac{9}{2}+\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right| \\
& =\log \left|x-\frac{9}{2}+\sqrt{x^2-9 x+20}\right| \ldots . . . . .(v)
\end{aligned}
$
Putting values of $\mathrm{I}_1$ and $\mathrm{I}_2$ in eq. (iii),
$
\mathrm{I}=6 \sqrt{x^2-9 x+20}+34 \log \left|x-\frac{9}{2}+\sqrt{x^2-9 x+20}\right|+c
$
Ex 7.4 Question 20.
$\frac{x+2}{\sqrt{4 x-x^2}}$
Answer.
Let $\mathrm{I}=\int \frac{x+2}{\sqrt{4 x-x^2}} d x$
$
\begin{aligned}
& \text { Let Linear }=\mathrm{A} \frac{d}{d x} \text { (Quadratic) }+\mathrm{B} \\
& \Rightarrow x+2=\mathrm{A} \frac{d}{d x}\left(4 x-x^2\right)+\mathrm{B} \\
& \Rightarrow x+2=\mathrm{A}(4-2 x)+\mathrm{B} \ldots \ldots . \text { (ii) } \\
& \Rightarrow x+2=4 \mathrm{~A}-2 \mathrm{~A} x+\mathrm{B}
\end{aligned}
$
Comparing coefficients of $x,-2 \mathrm{~A}=1 \Rightarrow \mathrm{A}=\frac{-1}{2}$
Comparing constants, $4 \mathrm{~A}+\mathrm{B}=2$
On solving, we get $\mathrm{A}=\frac{-1}{2}, \mathrm{~B}=4$
Putting the values of $A$ and $B$ in eq. (ii),
$
x+2=\frac{-1}{2}(4-2 x)+4
$
Putting this value of $x+2$ in eq. (i),
$
\begin{aligned}
\mathrm{I} & =\int \frac{\frac{-1}{2}(4-2 x)+4}{\sqrt{4 x-x^2}} d x \\
\mathrm{I} & =\frac{-1}{2} \int \frac{4-2 x}{\sqrt{4 x-x^2}} d x+4 \int \frac{1}{\sqrt{4 x-x^2}} d x \\
& \Rightarrow \mathrm{I}=\frac{-1}{2} \mathrm{I}_1+4 \mathrm{I}_2 \ldots \ldots \ldots . \text { (iii) }
\end{aligned}
$
$\begin{aligned}
& \text { Now } \mathrm{I}_1=\int \frac{4-2 x}{\sqrt{4 x-x^2}} d x \\
& \text { Putting } 4 x-x^2=t \\
& \Rightarrow 4-2 x=\frac{d t}{d x} \\
& \Rightarrow(4-2 x) d x=d t \\
& \therefore \mathrm{I}_1=\int \frac{d t}{\sqrt{t}}=\int t^{\frac{-1}{2}} d t \\
& =\frac{t^{1 / 2}}{1 / 2}=2 \sqrt{t} \\
& =2 \sqrt{4 x-x^2} \ldots \ldots . .(\mathrm{iv})
\end{aligned}$
$
\begin{aligned}
& \text { Again } \mathrm{I}_2=\int \frac{1}{\sqrt{4 x-x^2}} d x \\
& =\int \frac{1}{\sqrt{-x^2+4 x}} d x \\
& =\int \frac{1}{\sqrt{-\left(x^2-4 x\right)}} d x \\
& =\int \frac{1}{\sqrt{-\left(x^2-4 x+4-4\right)}} d x \\
& =\int \frac{1}{\sqrt{-\left[(x-2)^2-(2)^2\right]}} d x \\
& =\int \frac{1}{\sqrt{(2)^2-(x-2)^2}} d x \\
& =\sin ^{-1} \frac{x-2}{2} \ldots \ldots \ldots .(\mathrm{v})
\end{aligned}
$
Putting values of $I_1$ and $I_2$ in eq. (iii),
$\mathrm{I}=-\sqrt{4 x-x^2}+4 \sin ^{-1} \frac{x-2}{2}+c$
Ex 7.4 Question 21.
$\frac{x+2}{\sqrt{x^2+2 x+3}}$
Answer.
Let $\mathrm{I}=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x$
Let Linear $=\mathrm{A} \frac{d}{d x}($ Quadratic $)+\mathrm{B}$
$\Rightarrow x+2=A \frac{d}{d x}\left(x^2+2 x+3\right)+B$
$\Rightarrow x+2=A(2 x+2)+B$
$\Rightarrow x+2=2 \mathrm{~A} x+2 \mathrm{~A}+\mathrm{B}$
Comparing coefficients of $x, 2 \mathrm{~A}=1 \Rightarrow \mathrm{A}=\frac{1}{2}$
Comparing constants, $2 \mathrm{~A}+\mathrm{B}=2$
On solving, we get $\mathrm{A}=\frac{1}{2}, \mathrm{~B}=1$
Putting the values of $A$ and $B$ in eq. (ii),
$
x+2=\frac{1}{2}(2 x+2)+1
$
Putting this value of $x+2$ in eq. (i),
$
I=\int \frac{\frac{1}{2}(2 x+2)+1}{\sqrt{x^2+2 x+3}} d x
$
$\begin{aligned}
& \mathrm{I}=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^2+2 x+3}} d x+\int \frac{1}{\sqrt{x^2+2 x+3}} d x \\
& \Rightarrow \mathrm{I}=\frac{1}{2} \mathrm{I}_1+\mathrm{I}_2 \ldots \ldots \ldots . \text { (iii) } \\
& \text { Now } \mathrm{I}_1=\int \frac{2 x+2}{\sqrt{x^2+2 x+3}} d x \\
& \text { Putting } x^2+2 x+3=t \\
& \Rightarrow 2 x+2=\frac{d t}{d x} \\
& \Rightarrow(2 x+2) d x=d t \\
& \therefore \mathrm{I}_1=\int \frac{d t}{\sqrt{t}}=\int t^{\frac{-1}{2}} d t
\end{aligned}$
$\begin{aligned}
& =\frac{t^{1 / 2}}{1 / 2}=2 \sqrt{t} \\
& =2 \sqrt{x^2+2 x+3} \ldots \ldots \ldots . . \text { (iv) } \\
& \text { Again } \mathrm{I}_2=\int \frac{1}{\sqrt{x^2+2 x+3}} d x \\
& =\int \frac{1}{\sqrt{x^2+2 x+1+2}} \\
& =\int \frac{1}{\sqrt{(x+1)^2+(\sqrt{2})^2}} d x \\
& =\log \left|x+1+\sqrt{(x+1)^2+(\sqrt{2})^2}\right|
\end{aligned}$
$
=\log \left|x+1+\sqrt{x^2+2 x+3}\right|
$
Putting values of $\mathrm{I}_1$ and $\mathrm{I}_2$ in eq. (iii),
$
I=\sqrt{x^2+2 x+3}+\log \left|x+1+\sqrt{x^2+2 x+3}\right|+c
$
Ex 7.4 Question 22.
$\frac{x+3}{x^2-2 x-5}$
Answer.
Let $\mathrm{I}=\int \frac{x+3}{x^2-2 x-5} d x$
Let Linear $=\mathrm{A} \frac{d}{d x}($ Quadratic $)+\mathrm{B}$
$\Rightarrow x+3=\mathrm{A} \frac{d}{d x}\left(x^2-2 x-5\right)+\mathrm{B}$
$\Rightarrow x+3=\mathrm{A}(2 x-2)+\mathrm{B}$
$\Rightarrow x+3=2 \mathrm{~A} x-2 \mathrm{~A}+\mathrm{B}$
Comparing coefficients of $x, 2 \mathrm{~A}=1$
$\Rightarrow A=\frac{1}{2}$
Comparing constants, $-2 \mathrm{~A}+\mathrm{B}=3$
On solving, we get $\mathrm{A}=\frac{1}{2}, \mathrm{~B}=4$
Putting the values of A and B in eq. (ii), $x+3=\frac{1}{2}(2 x-2)+4$
Putting this value of $x+3$ in eq. (i),
$
\begin{aligned}
& \mathrm{I}=\int \frac{\frac{1}{2}(2 x+2)+4}{x^2-2 x-5} d x \\
& \mathrm{I}=\frac{1}{2} \int \frac{2 x-2}{x^2-2 x-5} d x+4 \int \frac{1}{x^2-2 x-5} d x \\
& \Rightarrow \mathrm{I}=\square
\end{aligned}
$
....(iii)
Now $\mathrm{I}_1=\int \frac{2 x-2}{x^2-2 x-5} d x$
Putting $x^2-2 x-5=t$
$
\begin{aligned}
& \Rightarrow 2 x-2=\frac{d t}{d x} \\
& \Rightarrow(2 x-2) d x=d t \\
& \therefore I_1=\int \frac{d t}{t}=\log |t| \\
& =\log \left|x^2-2 x-5\right| \ldots \ldots . .
\end{aligned}
$
$
\begin{aligned}
& \text { Again } \mathrm{I}_2=\int \frac{1}{x^2-2 x-5} d x \\
& =\int \frac{1}{\sqrt{x^2-2 x+1-1-5}} \\
& =\int \frac{1}{\sqrt{(x-1)^2+(\sqrt{6})^2}} d x \\
& =\frac{1}{2 \sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right| \ldots \ldots \ldots
\end{aligned}
$
Putting values of $I_1$ and $I_2$ in eq. (iii),
$
\mathrm{I}=\frac{1}{2} \log \left|x^2-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+c
$
Ex 7.4 Question 23.
$\frac{5 x+3}{\sqrt{x^2+4 x+10}}$
Answer.
Let $\mathrm{I}=\int \frac{5 x+3}{\sqrt{x^2+4 x+10}} d x$ ..(i)
Let Linear $=\mathrm{A} \frac{d}{d x}$ (Quadratic) + B
$\Rightarrow 5 x+3=\mathrm{A} \frac{d}{d x}\left(x^2+4 x+10\right)+\mathrm{B}$
$\Rightarrow 5 x+3=A(2 x+4)+B$
$\Rightarrow 5 x+3=2 A x+4 A+B$
Comparing coefficients of $x, 2 \mathrm{~A}=5 \Rightarrow \mathrm{A}=\frac{5}{2}$
Comparing constants, $4 \mathrm{~A}+\mathrm{B}=3$
On solving, we get $\mathrm{A}=\frac{1}{2}, \mathrm{~B}=-7$
Putting the values of A and B in eq. (ii),
$
5 x+3=\frac{5}{2}(2 x+4)-7
$
Putting this value of $5 x+3$ in eq. (i),
$\begin{aligned}
& \mathrm{I}=\int \frac{\frac{5}{2}(2 x+4)-7}{\sqrt{x^2+4 x+10}} d x \\
& \mathrm{I}=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^2+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^2+4 x+10}} d x \\
& \Rightarrow \mathrm{I}=\frac{5}{2} I_1-7 I_2 \ldots \\
& \text { Now } \mathrm{I}_1=\int \frac{2 x+4}{\sqrt{x^2+4 x+10}} d x \\
& \text { Putting } x^2+4 x+10=t \\
& \Rightarrow 2 x+4=\frac{d t}{d x} \\
& \Rightarrow(2 x+4) d x=d t \\
& \therefore \mathrm{I}_1=\int \frac{d t}{\sqrt{t}} \\
& =\int t^{\frac{-1}{2}} d t \\
& =\frac{t^{1 / 2}}{1 / 2}=2 \sqrt{t} \\
& =2 \sqrt{x^2+4 x+10} \\
& \text { Again } \mathrm{I}_2=\int \frac{1}{\sqrt{x^2+4 x+10}} d x \\
&
\end{aligned}$
$=\int \frac{1}{\sqrt{x^2+4 x+4+6}}$
$
\begin{aligned}
& =\int \frac{1}{\sqrt{(x+2)^2+(\sqrt{6})^2}} d x \\
& =\log \left|x+2+\sqrt{(x+2)^2+(\sqrt{6})^2}\right| \\
& =\log \left|x+2+\sqrt{x^2+4 x+10}\right| \ldots \ldots \ldots . .(v)
\end{aligned}
$
Putting values of $\mathrm{I}_1$ and $\mathrm{I}_2$ in eq. (iii),
$
\mathrm{I}=5 \sqrt{x^2+4 x+10}-7 \log \left|x+2+\sqrt{x^2+4 x+10}\right|+c
$
Ex 7.4 Question 24.
$\int \frac{d x}{x^2+2 x+2}$ equals
(A) $x \tan ^{-1}(x+1)+\mathrm{C}$
(B) $\tan ^{-1}(x+1)+C$
(C) $(x+1) \tan ^{-1} x+C$
(D) $\tan ^{-1} x+\mathrm{C}$
Answer.
$\int \frac{d x}{x^2+2 x+2}$
$
\begin{aligned}
& =\int \frac{1}{x^2+2 x+1+1} d x \\
& =\int \frac{1}{(x+1)^2+(1)^2} d x \\
& =\tan ^{-1}(x+1)+C
\end{aligned}
$
Ex 7.4 Question 25.
$\int \frac{d x}{\sqrt{9 x-4 x^2}}$ equals
(A) $\frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C$
(B) $\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+\mathrm{C}$
(C) $\frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C$
(D) $\frac{1}{2} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+\mathrm{C}$
Answer.
Let $\mathrm{I}=\int \frac{d x}{\sqrt{9 x-4 x^2}}$
$
\begin{aligned}
& =\int \frac{1}{\sqrt{-4 x^2+9 x}} d x \\
& =\int \frac{1}{\left.\sqrt{-4\left(x^2-\frac{9}{4} x\right.}\right)} d x \\
& =\int \frac{1}{\sqrt{-4\left(x^2-\frac{9}{4} x+\left(\frac{9}{8}\right)^2-\left(\frac{9}{8}\right)^2\right)}} d x \\
& =\int \frac{1}{\sqrt{-4\left[\left(x-\frac{9}{8}\right)^2+\left(\frac{9}{8}\right)^2\right]}} d x \\
& =\int \frac{1}{\sqrt{4\left[\left(\frac{9}{8}\right)^2-\left(x-\frac{9}{8}\right)^2\right]}} d x
\end{aligned}
$
$=\frac{1}{2} \int \frac{1}{\sqrt{\left[\left(\frac{9}{8}\right)^2-\left(x-\frac{9}{8}\right)^2\right]}} d x$
$
\begin{aligned}
& =\frac{1}{2} \sin ^{-1} \frac{x-\frac{9}{8}}{\frac{9}{8}}+C \\
& =\frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+\mathrm{C}
\end{aligned}
$
Therefore, option (B) is correct.
