Exercise 11.11 - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.11
Integrate the following with respect to $\mathrm{x}$ :
Question 1.
(i) $\frac{2 x-3}{x^2+4 x-12}$
(ii) $\frac{5 x-2}{2+2 x+x^2}$
(iii) $\frac{3 x+1}{2 x^2-2 x+3}$
Solution:
Let
$
\begin{aligned}
\mathrm{I} & =\int \frac{2 x-3}{x^2+4 x-12} d x \\
2 x-3 & =\mathrm{A} \frac{d}{d x}\left(x^2+4 x-12\right)+\mathrm{B} \\
2 x-3 & =\mathrm{A}(2 x+4)+\mathrm{B}
\end{aligned}
$
Comparing the coefficients of like terms, we get,
$
\begin{aligned}
2 \mathrm{~A} & =2 \\
4 \mathrm{~A}+\mathrm{B} & =-3 \\
4+\mathrm{B} & =-3 \Rightarrow \mathrm{B}=-7 \\
\mathrm{I} & =\int \frac{(2 x+4)-7}{x^2+4 x-12} d x \\
& =\int \frac{2 x+4}{x^2+4 x-12} d x-7 \int \frac{1}{x^2+4 x-12} d x \\
\mathrm{I} & =\log \left|x^2+4 x-12\right|-7 \mathrm{I}_1 \\
& \mathrm{I}_1=\int \frac{1}{x^2+4 x-12} d x \\
& =\int \frac{1}{(x+2)^2-4-12} d x \\
& =\int \frac{1}{(x+2)^2-(4)^2} d x
\end{aligned}
$

$
\begin{aligned}
& =\frac{1}{2(4)} \log \left|\frac{x+2-4}{x+2+4}\right| \\
\mathrm{I}_1 & =\frac{1}{8} \log \left|\frac{x-2}{x+6}\right|
\end{aligned}
$
Substituting $I_1$, value in (1) we get,
$
\therefore \quad \mathrm{I}=\log \left|x^2+4 x-12\right|-\frac{7}{8} \log \left|\frac{x-2}{x+6}\right|+c
$

(ii) Let
$
\begin{aligned}
\mathrm{I} & =\int \frac{5 x-2}{x^2+2 x+2} d x \\
5 x-2 & =\mathrm{A} \frac{d}{d x}\left(x^2+2 x+2\right)+\mathrm{B} \\
5 x-2 & =\mathrm{A}(2 x+2)+\mathrm{B}
\end{aligned}
$
Comparing the coefficient of like term we get
$
\begin{aligned}
& 2 \mathrm{~A}=5 \\
& \mathrm{~A}=\frac{5}{2} \\
& 2 \mathrm{~A}+\mathrm{B}=-2 \\
& 5+\mathrm{B}=-2 \\
& \mathrm{~B}=-7 \\
& I=\int \frac{\frac{5}{2}(2 x+2)-7}{x^2+2 x+2} d x \\
& =\frac{5}{2} \int \frac{2 x+2}{x^2+2 x+2} d x-7 \int \frac{1}{x^2+2 x+2} d x \\
& I=\frac{5}{2} \log \left|x^2+2 x+2\right|-7 I_1 \\
& I_1=\int \frac{1}{x^2+2 x+2} d x=\int \frac{1}{(x+1)^2-1+2} d x \\
& =\int \frac{1}{(x+1)^2+(1)^2} d x \\
&
\end{aligned}
$

$
\begin{aligned}
\int \frac{1}{x^2+a^2} d x & =\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c \\
& =\frac{1}{1} \tan ^{-1}\left(\frac{x+1}{1}\right) \\
\mathrm{I}_1 & =\tan ^{-1}(x+1)
\end{aligned}
$
Substituting $\mathrm{I}_1$ value in (1) we get,
$
\therefore \quad \mathrm{I}=\frac{5}{2} \log \left|x^2+2 x+2\right|-7 \tan ^{-1}(x+1)+c
$

(iii) Let
$
\begin{aligned}
\mathrm{I} & =\int \frac{3 x+1}{2 x^2-2 x+3} d x \\
3 x+1 & =\mathrm{A} \frac{d}{d x}\left(2 x^2-2 x+3\right)+\mathrm{B} \\
3 x+1 & =\mathrm{A}(4 x-2)+\mathrm{B}
\end{aligned}
$
Comparing the coefficients of like terms, we get
$
\begin{aligned}
4 \mathrm{~A} & \left.=3 \Rightarrow \mathrm{A}=\frac{3}{4}\right] \\
-2 \mathrm{~A}+\mathrm{B} & =1=-2\left(\frac{3}{4}\right)+\mathrm{B}=1 \\
\mathrm{~B} & \left.=1+\frac{3}{2} \Rightarrow \mathrm{B}=\frac{5}{2}\right] \\
\mathrm{I} & =\int \frac{3}{2 x^2-2 x+3}(4 x-2)+\frac{5}{2} \\
\mathrm{I} & =\frac{3}{4} \log \left|2 x^2-2 x+3\right|+\frac{5}{2} I_1 \\
& =\frac{3}{4} \int \frac{4 x-2}{2 x^2-2 x+3} d x+\frac{5}{2} \int \frac{1}{2 x^2-2 x+3} d x \\
\mathrm{I}_1 & =\int \frac{1}{2 x^2-2 x+3} d x \\
2 x^2-2 x+3 & =2\left[x^2-x+\frac{3}{2}\right] \\
& =2\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+\frac{3}{2}\right]=2\left[\left(x-\frac{1}{2}\right)^2+\frac{5}{4}\right]
\end{aligned}
$

$
\begin{aligned}
& =2\left[\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{5}}{2}\right)^2\right] \\
I_1 & =\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{5}}{2}\right)^2} d x \\
I_1 & =\frac{1}{2}\left[\frac{1}{\sqrt{5} / 2} \tan ^{-1}\left(\frac{x-\frac{1}{2}}{\sqrt{\frac{5}{2}}}\right)\right]=\frac{1}{2}\left[\frac{2}{\sqrt{5}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)\right] \\
I_1 & =\frac{1}{\sqrt{5}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)
\end{aligned}
$
Substituting $I_1$ value in (1) we get,
$
\begin{gathered}
\therefore \quad \mathrm{I}=\frac{3}{4} \log \left|2 x^2-2 x+3\right|+\frac{5}{2}\left[\frac{1}{\sqrt{5}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)\right]+c \\
\mathrm{I}=\frac{3}{4} \log \left|2 x^2-2 x+3\right|+\frac{\sqrt{5}}{2} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)+c
\end{gathered}
$
Question 2 .
(i) $\frac{2 x+1}{\sqrt{9+4 x-x^2}}$
(ii) $\frac{x+2}{\sqrt{x^2-1}}$
(iii) $\frac{2 x+3}{\sqrt{x^2+4 x+1}}$
Solution:

(i) Let
$
\begin{aligned}
\mathrm{I} & =\int \frac{2 x+1}{\sqrt{9+4 x-x^2}} d x \\
2 x+1 & =A \frac{d}{d x}\left(9+4 x-x^2\right)+\mathrm{B} \\
2 x+1 & =\mathrm{A}(4-2 x)+\mathrm{B}
\end{aligned}
$
Comparing the coefficients of like terms, we get
$
\begin{aligned}
-2 \mathrm{~A} & =2 \Rightarrow \mathrm{A}=-1 \\
4 \mathrm{~A}+\mathrm{B} & =1 \\
-4+\mathrm{B} & =1 \Rightarrow \mathrm{B}=5 \\
\mathrm{I} & =\int \frac{-(4-2 x)}{\sqrt{9+4 x-x^2}} d x+5 \int \frac{1}{\sqrt{9+4 x-x^2}} d x \\
\mathrm{I} & =-2 \sqrt{9+4 x-x^2}+5 I_1
\end{aligned}
$
Consider,
$
\begin{aligned}
I_1 & =\int \frac{1}{\sqrt{9+4 x-x^2}} d x \\
-x^2+4 x+9 & =-\left[x^2-4 x-9\right] . \\
& =-\left[(x-2)^2-4-9\right] \\
& =-\left[(x-2)^2-(\sqrt{13})^2\right] \\
& =(\sqrt{13})^2-(x-2)^2 \\
\quad & I_1=\int \frac{1}{\sqrt{(\sqrt{13})^2-(x-2)^2}} d x=\sin ^{-1}\left(\frac{x-2}{\sqrt{13}}\right)
\end{aligned}
$
Substituting $I_1$ value in (1) we get,
$
\therefore \quad \mathrm{I}=-2 \sqrt{9+4 x-x^2}+5 \sin ^{-1}\left(\frac{x-2}{\sqrt{13}}\right)+c
$

(ii) Sol. Let
$
\begin{aligned}
\mathrm{I} & =\int \frac{x+2}{\sqrt{x^2-1}} d x \\
x+2 & =\mathrm{A} \frac{d}{d x}\left(x^2-1\right)+\mathrm{B} \\
x+2 & =\mathrm{A}(2 x)+\mathrm{B}
\end{aligned}
$
Comparing the coefficients of like terms, we get
$
\begin{aligned}
2 \mathrm{~A} & =1, \quad \mathrm{~B}=2 \\
\mathrm{~A} & \left.=\frac{1}{2}\right] \\
\mathrm{I} & =\frac{1}{2} \int \frac{(2 x)}{\sqrt{x^2-1}} d x+2 \int \frac{1}{\sqrt{x^2-1}} d x \\
\mathrm{I} & =\frac{1}{2}\left[2 \sqrt{x^2-1}\right]+2 \mathrm{I}_1 \\
\mathrm{I} & =\sqrt{x^2-1}+2 \mathrm{I}_1 \\
\text { Consider, } \quad \mathrm{I}_1=\int \frac{1}{\sqrt{x^2-1}} d x & =\log \left|x+\sqrt{x^2-1}\right|+c
\end{aligned}
$
Substituting $I_1$ value in (1) we get,
$
\therefore \quad \mathrm{I}=\sqrt{x^2-1}+2 \log \left|x+\sqrt{x^2}-1\right|+c
$

(iii) Sol. Let
$
\begin{aligned}
\mathrm{I} & =\int \frac{2 x+3}{\sqrt{x^2+4 x+1}} d x \\
2 x+3 & =A \frac{d}{d x}\left(x^2+4 x+1\right)+B \\
2 x+3 & =A(2 x+4)+B
\end{aligned}
$
Comparing the coefficients of like terms, we get,
$
\text { Consider, } x^2+4 x+1
$
$
\begin{aligned}
& \begin{array}{l|l}
2 \mathrm{~A}=2 & 4 \mathrm{~A}+\mathrm{B}=3 \\
\mathrm{~A}=1 & 4+\mathrm{B}=3 \\
& \mathrm{~B}=-1
\end{array} \\
& I=\int \frac{2 x+4}{\sqrt{x^2+4 x+1}} d x-\int \frac{1}{\sqrt{x^2+4 x+1}} d x \\
& I=2 \sqrt{x^2+4 x+1}-I_1 \\
& =(x+2)^2-4+1 \\
& =(x+2)^2-(\sqrt{3})^2 \\
&
\end{aligned}
$

$\begin{aligned}
I_1 & =\int \frac{1}{\sqrt{(x+2)^2-(\sqrt{3})^2}} d x \\
& =\log \left|(x+2)+\sqrt{(x+2)^2-(\sqrt{3})^2}\right|+c \\
\mathrm{I}_1 & =\log \left|(x+2)+\sqrt{x^2+4 x+1}\right|+c \\
\mathrm{I} & =2 \sqrt{x^2+4 x+1}-\log \left|(x+2)+\sqrt{x^2+4 x+1}\right|+c
\end{aligned}$