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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
$\frac{2 x-1}{2 x^2+x+3}$
Solution:
Let $2 x-1=\mathrm{A} \frac{d}{d x}\left(2 x^2+x+3\right)+\mathrm{B}$
$
2 x-1=\mathrm{A}(4 x+1)+\mathrm{B}
$
Comparing the coefficients like terms, we get
$
\begin{aligned}
& 2=4 \mathrm{~A} \Rightarrow \mathrm{A}=\frac{1}{2} \\
& -1=\mathrm{A}+\mathrm{B} \\
& \frac{1}{2}+B=-1 \Rightarrow B=\frac{-3}{2} \\
& \therefore \quad 2 x-1=\frac{1}{2}(4 x+1)+\left(\frac{-3}{2}\right) \\
& \int \frac{2 x-1}{2 x^2+x+3} d x=\int \frac{\frac{1}{2}(4 x+1)+\left(\frac{-3}{2}\right)}{2 x^2+x+3} d x \\
& =\frac{1}{2} \int \frac{4 x+1}{2 x^2+x+3} d x-\frac{3}{2} \int \frac{d x}{2 x^2+x+3} \\
& =\frac{1}{2} \log \left(2 x^2+x+3\right)-\frac{3}{2} I_1 \\
& I_1=\int \frac{d x}{2 x^2+x+3}=2 \int \frac{d x}{4 x^2+2 x+6} \\
& 4 x^2+2 x+6=(2 x)^2+2(2 x)\left(\frac{1}{2}\right)+6+\frac{1}{4}-\frac{1}{4} \\
&
\end{aligned}
$

$
\begin{aligned}
& =\left(2 x+\frac{1}{2}\right)^2+\frac{23}{4}=\left(2 x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{23}}{2}\right)^2 \\
& \therefore \quad \mathrm{I}_1=\int \frac{2 d x}{\left(2 x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{23}}{2}\right)^2} \\
& =2\left\{\frac{\frac{1}{\sqrt{23}}}{2}\right\} \tan ^{-1}\left(\frac{2 x+1 / 2}{\frac{\sqrt{23}}{2}}\right) \\
& =\frac{4}{\sqrt{23}} \tan ^{-1}\left(\frac{4 x+1}{\frac{\sqrt{23}}{2}}\right)=\frac{2}{\sqrt{23}} \tan ^{-1}\left(\frac{4 x+1}{\sqrt{23}}\right) \\
&
\end{aligned}
$
Substituting $I_1$ value in (1) we get
$
\begin{aligned}
& \frac{1}{2} \log \left(2 x^2+x+3\right) \mathrm{I}=1-\frac{3}{2}\left(\frac{2}{\sqrt{23}} \tan ^{-1} \frac{4 x+1}{\sqrt{23}}\right) \\
&=\frac{1}{2} \log \left(2 x^2+x+3\right)-\frac{3}{\sqrt{23}} \tan ^{-1}\left(\frac{4 x+1}{\sqrt{23}}\right)+c
\end{aligned}
$
Question 2.
$
\frac{4 x+1}{x^2+3 x+1}
$
Solution:

Let
$
\begin{aligned}
& 4 x+1=\mathrm{A} \frac{d}{d x}\left(x^2+3 x+1\right)+\mathrm{B} \\
& 4 x+1=\mathrm{A}(2 x+3)+\mathrm{B}
\end{aligned}
$
Comparing the coefficients like terms, we get,
$
\begin{aligned}
& 4=2 A \Rightarrow A=2 \\
& 1=3 \mathrm{~A}+\mathrm{B} \\
& 3(2)+\mathrm{B}=1 \\
& \mathrm{~B}=1-6=-5 \\
& 4 x+1=2(2 x+3)-5 \\
& \therefore \quad \int \frac{4 x+1}{x^2+3 x+1} d x=\int \frac{2(2 x+3)-5}{x^2+3 x+1} \\
& =2 \int \frac{2 x+3}{x^2+3 x+1} d x-5 \int \frac{d x}{x^2+3 x+1} \\
& =2 \log \left(x^2+3 x+1\right)-5 I_1 \\
& \mathrm{I}_1=\int \frac{d x}{x^2+3 x+1} \\
& x^2+3 x+1=x^2+2\left(\frac{3}{2}\right) x+1+\frac{9}{4}-\frac{9}{4} \\
& =\left(x+\frac{3}{2}\right)^2-\frac{5}{4} \\
& =\left(x+\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2 \\
& \therefore \quad \mathrm{I}_1=\int \frac{d x}{\left(x+\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2} \\
& =\frac{1}{2 \sqrt{5} / 2} \log \left(\frac{x+3 / 2-\sqrt{5} / 2}{x+3 \sqrt{2}+\sqrt{5} / 2}\right) \\
& =\frac{1}{\sqrt{5}} \log \left(\frac{2 x+3-\sqrt{5}}{2 x+3+\sqrt{5}}\right) \\
&
\end{aligned}
$

Substituting $I_1$ value in (1) we get,
$
\begin{aligned}
& I=2 \log \left(x^2+3 x+1\right) \frac{-5}{\sqrt{5}} \log \left(\frac{2 x+3-\sqrt{5}}{2 x+3+\sqrt{5}}\right) \\
&=2 \log \left(x^2+3 x+1\right)-\sqrt{5} \log \left(\frac{2 x+3-\sqrt{5}}{2 x+3+\sqrt{5}}\right)+c
\end{aligned}
$

Let
$
\begin{aligned}
& 4 x+1=\mathrm{A} \frac{d}{d x}\left(x^2+3 x+1\right)+\mathrm{B} \\
& 4 x+1=\mathrm{A}(2 x+3)+\mathrm{B}
\end{aligned}
$
Comparing the coefficients like terms, we get,
$
\begin{aligned}
& 4=2 \mathrm{~A} \Rightarrow \mathrm{A}=2 \\
& 1=3 \mathrm{~A}+\mathrm{B} \\
& 3(2)+\mathrm{B}=1 \\
& \mathrm{~B}=1-6=-5 \\
& 4 x+1=2(2 x+3)-5 \\
& \therefore \quad \int \frac{4 x+1}{x^2+3 x+1} d x=\int \frac{2(2 x+3)-5}{x^2+3 x+1} \\
& =2 \int \frac{2 x+3}{x^2+3 x+1} d x-5 \int \frac{d x}{x^2+3 x+1} \\
& =2 \log \left(x^2+3 x+1\right)-5 \mathrm{I}_1 \\
& \mathrm{I}_1=\int \frac{d x}{x^2+3 x+1} \\
& x^2+3 x+1=x^2+2\left(\frac{3}{2}\right) x+1+\frac{9}{4}-\frac{9}{4} \\
& =\left(x+\frac{3}{2}\right)^2-\frac{5}{4} \\
& =\left(x+\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2 \\
& \therefore \quad \mathrm{I}_1=\int \frac{d x}{\left(x+\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2} \\
& =\frac{1}{2 \sqrt{5} / 2} \log \left(\frac{x+3 / 2-\sqrt{5} / 2}{x+3 \sqrt{2}+\sqrt{5} / 2}\right) \\
& =\frac{1}{\sqrt{5}} \log \left(\frac{2 x+3-\sqrt{5}}{2 x+3+\sqrt{5}}\right) \\
&
\end{aligned}
$
Substituting $I_1$ value in (1) we get,

$\begin{aligned}
& I=2 \log \left(x^2+3 x+1\right) \frac{-5}{\sqrt{5}} \log \left(\frac{2 x+3-\sqrt{5}}{2 x+3+\sqrt{5}}\right) \\
& =2 \log \left(x^2+3 x+1\right)-\sqrt{5} \log \left(\frac{2 x+3-\sqrt{5}}{2 x+3+\sqrt{5}}\right)+c \\
&
\end{aligned}$

Question 3.
$\frac{2 x-3}{\sqrt{10-7 x-x^2}}$
Solution:
$
\begin{aligned}
& 2 x-3=A \frac{d}{d x}\left(10-7 x-x^2\right)+B \\
& 2 x-3=A(-7-2 x)+B
\end{aligned}
$
Comparing the coefficients like terms, we get,
$
\begin{aligned}
& 2=-2 \mathrm{~A} \Rightarrow \mathrm{A}=-1 \\
& -3=-7 \mathrm{~A}+\mathrm{B} \\
& -7(-1)+\mathrm{B}=-3 \\
& 7+B=-3 \\
& \mathrm{~B}=-3-7=-10 \\
& \therefore \quad 2 x-3=-1(-7-2 x)-10 \\
& \therefore \quad \int \frac{2 x-3}{\sqrt{10-7 x-x^2}} d x=\int \frac{-1(-7-2 x)-10}{\sqrt{10-7 x-x^2}} d x \\
& =-\int \frac{(-7-2 x) d x}{\sqrt{10-7 x-x^2}}-10 \int \frac{d x}{\sqrt{10-7 x-x^2}} \\
& =-\left[2 \sqrt{10-7 x-x^2}\right]-10 \mathrm{I}_1 \\
& I_1=\int \frac{d x}{\sqrt{10-7 x-x^2}} \\
& 10-7 x-x^2=-\left(x^2+7 x-10\right) \\
& =-\left[x^2+2\left(\frac{7}{2}\right) x-10+\frac{49}{4}-\frac{49}{4}\right] \\
& =-\left[\left(x+\frac{7}{2}\right)^2-\frac{89}{4}\right]=\left(\frac{\sqrt{89}}{2}\right)^2-\left(x+\frac{7}{2}\right)^2 \\
& \mathrm{I}_1=\int \frac{d x}{\sqrt{\left(\frac{\sqrt{89}}{2}\right)^2-\left(x+\frac{7}{2}\right)^2}} \\
& =\sin ^{-1}\left(\frac{x+\frac{7}{2}}{\frac{\sqrt{89}}{2}}\right)=\sin ^{-1}\left(\frac{2 x+7}{\sqrt{89}}\right) \\
&
\end{aligned}
$

Substituting $\mathrm{I}_1$ value in (1) we get,
$
I=-2 \sqrt{10-7 x-x^2}-10 \sin ^{-1}\left(\frac{2 x+7}{\sqrt{89}}\right)+c
$
Question 4.
$
\frac{6 x+7}{\sqrt{(x-4)(x-5)}}
$
Solution:

$
\frac{6 x+7}{\sqrt{(x-4)(x-5)}}=\frac{6 x+7}{\sqrt{x^2-9 x+20}}
$
Let
$
\begin{aligned}
& 6 x+7=\mathrm{A} \frac{d}{d x}\left(x^2-9 x+20\right)+\mathrm{B} \\
& 6 x+7=\mathrm{A}(2 x-9)+\mathrm{B}
\end{aligned}
$
Comparing the coefficients like terms, we get,
$
\begin{aligned}
6 & =2 \mathrm{~A} \Rightarrow \mathrm{A}=3 \\
7 & =-9 \mathrm{~A}+\mathrm{B} \\
-9(3)+\mathrm{B} & =7 \\
\mathrm{~B} & =7+27=34 \\
\mathrm{I} & =\int \frac{6 x+7}{\sqrt{x^2-9 x+20}} d x \\
& =\int \frac{3(2 x-9)+34}{\sqrt{x^2-9 x+20}} d x \\
& =3 \int \frac{2 x-9}{\sqrt{x^2-9 x+20}} d x+34 \int \frac{d x}{\sqrt{x^2-9 x+20}} \\
& =3\left\{2 \sqrt{\left.x^2-9 x+20\right\}}+34 \mathrm{I}_1\right. \\
\mathrm{I}_1 & =\int \frac{d x}{\sqrt{x^2-9 x+20}} \\
& =34 \mathrm{I}_1
\end{aligned}
$

$
\begin{aligned}
x^2-9 x+20 & =x^2-2\left(\frac{9}{2}\right) x+20+\frac{81}{4}-\frac{81}{4} \\
& =\left(x-\frac{9}{2}\right)^2-\frac{1}{4} \\
& =\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2 \\
I_1 & =\int \frac{d x}{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\
& =\log \left(x-\frac{9}{2}\right)+\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2} \\
& =\log \left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}
\end{aligned}
$
Substituting $I_1$ value in (1) we get
$
I=6 \sqrt{x^2-9 x+20}+34\left[\log \left(x-\frac{9}{2}\right)+\sqrt{x^2-9 x+20}\right]+c
$