Exercise 2.2 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $2.2$
Integrate the following with respect to $\mathrm{x}$.
Question 1.
$
\int\left(\sqrt{2 x}-\frac{1}{\sqrt{2 x}}\right)^2
$
Solution:
$
\begin{aligned}
\int\left(\sqrt{2 x}-\frac{1}{\sqrt{2 x}}\right)^2 & =(\sqrt{2 x})^2-2(\sqrt{2 x})\left(\frac{1}{\sqrt{2 x}}\right)+\frac{1}{(\sqrt{2 x})^2} \\
& =2 x-2+\frac{1}{2 x}
\end{aligned}
$
So $\int\left(\sqrt{2 x}-\frac{1}{\sqrt{2 x}}\right)^2 d x=\int 2 x d x-\int 2 d x+\int \frac{1}{2 x} d x+c$
$
=x^2-2 x+\frac{1}{2} \log |x|+c
$
Question 2.
$\frac{x^4-x^2+2}{x-1}$
Solution:
$
\begin{aligned}
\frac{x^4-x^2+2}{x-1} & =\frac{x^2\left(x^2-1\right)+2}{x-1}=\frac{x^2\left(x^2-1\right)}{x-1}+\frac{2}{x-1} \\
& =\frac{x^2(x+1)(x-1)}{x-1}+\frac{2}{x-1} \\
& =x^3+x^2+\frac{2}{x-1}
\end{aligned}
$
So $\int \frac{x^4-x^2+2}{x-1} d x=\int x^3 d x+\int x^2 d x+\int \frac{2}{x-1} d x+c$
$
=\frac{x^4}{4}+\frac{x^3}{3}+2 \log |x-1|+c
$

Question 3.
$
\frac{x^3}{x+2}
$
Solution:
$
\begin{aligned}
\frac{x^3}{x+2}= & \frac{x^3+8-8}{x+2}=\frac{x^3+2^3}{x+2}-\frac{8}{x+2} \\
& \frac{(x+2)\left(x^2-2 x+4\right)}{x+2}-\frac{8}{x+2} \\
= & x^2-2 x+4-\frac{8}{x+2}
\end{aligned}
$
So
$
\begin{aligned}
\frac{x^3}{x+2} d x & =\int x^2 d x-\int 2 x d x+\int 4 d x-\int \frac{8}{x+2} d x+c \\
& =\frac{x^3}{3}-x^2+4 x-8 \log |x+2|+c
\end{aligned}
$
Question 4.
$
\frac{x^3+3 x^2-7 x+11}{x+5}
$
Solution:
We have to find the quotient by division

Question 5.
$
\frac{3 x+2}{(x-2)(x-3)}
$
Solution:
We use the partial fraction method to split the given function into two fractions and then
integrate.
$
\begin{aligned}
\frac{3 x+2}{(x-2)(x-3)} & =\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3} \\
3 x+2 & =\mathrm{A}(x-3)+\mathrm{B}(x-2)
\end{aligned}
$
Put $(x=3) \quad 11=\mathrm{B} \quad \Rightarrow \mathrm{B}=11$
Put $(x=2)$ in the identity
So $\quad \frac{3 x+2}{(x-2)(x-3)}=\frac{-8}{x-2}+\frac{11}{x-3}$
$
\begin{aligned}
3(2)+2 & =\mathrm{A}(2-3) \\
8 & =-\mathrm{A} \\
\Rightarrow \mathrm{A} & =-8
\end{aligned}
$
Thus $\int \frac{3 x+2}{(x-2)(x-3)} d x=-8 \int \frac{d x}{x-2}+11 \int \frac{d x}{x-3}+c$
$
\begin{aligned}
& =-8 \log |x-2|+11 \log |x-3|+c \\
\int \frac{3 x+2}{(x-2)(x-3)} d x & =11 \log |x-3|-8 \log |x-2|+c
\end{aligned}
$

Question 6.
$
\frac{4 x^2+2 x+6}{(x+1)^2(x-3)}
$
Solution:
By partial fractions,

$
\frac{4 x^2+2 x+6}{(x+1)^2(x-3)}=\frac{A}{x-3}+\frac{\mathrm{B}}{x+1}+\frac{\mathrm{C}}{(x+1)^2}
$
or $4 x^2+2 x+6=\mathrm{A}(x+1)^2+\mathrm{B}(x+1)(x-3)+\mathrm{C}(x-3)$.
Put $x=3$ on both sides of (1)
$
\begin{aligned}
4(9)+6+6 & =\mathrm{A}(3+1)^2 \\
48 & =16 \mathrm{~A} \\
\Rightarrow \quad \mathrm{A} & =3
\end{aligned}
$
Put $x=-1$ on both sides of (1)
$
\begin{aligned}
& 4(1)+2(-1)+6=C(-1-3) \\
& 8=-4 C \\
& \Rightarrow \quad \quad C=-2 \\
&
\end{aligned}
$
Put $x=0$ on both sides of (1)
$
\begin{aligned}
6 & =\mathrm{A}-3 \mathrm{~B}-3 \mathrm{C} \\
6 & =3-3 \mathrm{~B}-3(-2) \\
3 \quad \mathrm{~B} & =3-6+6 \\
\Rightarrow \quad \mathrm{B} & =1
\end{aligned}
$

$\begin{aligned}
\Rightarrow \quad \frac{4 x^2+2 x+6}{(x+1)^2(x-3)} & =\frac{3}{x-3}+\frac{1}{x+1}-\frac{2}{(x+1)^2} \\
\text { So } \int \frac{4 x^2+2 x+6}{(x+1)^2(x-3)} d x & =\int \frac{3}{x-3} d x+\int \frac{1}{x+1} d x-\int \frac{2}{(x+1)^2} d x \\
& =3 \log |x-3|+\log |x+1|-2 \int(x+1)^{-2} d x+c \\
& =3 \log |x-3|+\log |x+1|-\frac{2(x+1)^{-1}}{(-1)}+c \\
& =3 \log |x-3|+\log |x+1|+\frac{2}{x+1}+c
\end{aligned}$

Question 7.
$
\frac{3 x^2-2 x+5}{(x-1)\left(x^2+5\right)}
$
Solution:

$
\begin{aligned}
\frac{3 x^2-2 x+5}{(x-1)\left(x^2+5\right)} & =\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+5} \\
\Rightarrow \quad 3 x^2-2 x+5 & =\mathrm{A}\left(x^2+5\right)+(\mathrm{B} x+\mathrm{C})(x-1) .
\end{aligned}
$
Put $x=1$ on both sides of (1)
$
\begin{aligned}
3-2+5 & =A(1+5) \\
6 & =6 \mathrm{~A} \\
\Rightarrow \mathrm{A} & =1
\end{aligned}
$
Put $x=0$ on both sides of (1)
$
\begin{aligned}
5 & =5-\mathrm{C} \\
\mathrm{C} & =0
\end{aligned}
$
Put $x=2$ on both sides of (1)
$
\begin{aligned}
12-4+5 & =1(4+5)+(2 \mathrm{~B})(2-1) \\
13 & =9+2 \mathrm{~B} \\
2 \mathrm{~B} & =4 \\
\mathrm{~B} & =2
\end{aligned}
$
So $\frac{3 x^2-2 x+5}{(x-1)\left(x^2+5\right)}=\frac{1}{x-1}+\frac{2 x}{x^2+5}$
$
\Rightarrow \int \frac{3 x^2-2 x+5}{(x-1)\left(x^2+5\right)} d x=\int \frac{1}{x-1} d x+\int \frac{2 x}{x^2+5} d x+c
$
Now
$
\int \frac{2 x}{x^2+5}=\int \frac{d\left(x^2+5\right)}{x^2+5}=\log \left|x^2+5\right|
$
$
\begin{aligned}
\Rightarrow \quad \text { The solution is } & =\log |x-1|+\log \left|x^2+5\right|+c \\
& =\log \left|(x-1)\left(x^2+5\right)\right|+c \\
& =\log \left|x^3-x^2+5 x-5\right|+c
\end{aligned}
$

Question 8.
If $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{x}$ and $\mathrm{f}(1)=\frac{\pi}{4}$, then find $\mathrm{f}(\mathrm{x})$
Solution:
$
\begin{aligned}
f^{\prime}(x)=\frac{1}{x} \Rightarrow \int f^{\prime}(x) d x & =\int \frac{1}{x} d x \\
\Rightarrow f(x) & =\log |x|+c \\
\text { Using } f(1) & =\frac{\pi}{4}, \text { we get } \\
\frac{\pi}{4} & =\log 1+c \\
\Rightarrow c & =\frac{\pi}{4}(\because \log 1=0) \\
\text { Thus } f(x) & =\log |x|+\frac{\pi}{4}
\end{aligned}
$
Type: III
(i) $\int e^x d x=e^x+c$
(ii) $\int e^{a x+b} d x=\frac{1}{a} e^{a x+b}+c$
(iii) $\int a^x d x=\frac{1}{\log a} a^x+c, a>0$ and $a \neq 1$
(iv) $\int a^{m x+n} d x=\frac{1}{m \log a} a^{m x+n}+c, a>0$ and $a \neq 1$