Exercise 11.10 - 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.
(i) $\frac{1}{16-x^2}$
(ii) $\frac{1}{7-(4 x+1)^2}$
(iii) $\frac{1}{5-6 x-9 x^2}$
Solution:
(i) $\int \frac{d x}{16-x^2}=\int \frac{d x}{4^2-x^2}=\frac{1}{2(4)} \log \frac{4+x}{4-x}=\frac{1}{8} \log \frac{4+x}{4-x}+c$
(ii)
$
\begin{aligned}
I & =\int \frac{1}{7-(4 x+1)^2} d x=\int \frac{d x}{(\sqrt{7})^2-(4 x+1)^2} \\
& =\frac{1}{2 \sqrt{7}} \frac{\log \frac{\sqrt{7}+(4 x+1)}{\sqrt{7}-(4 x+1)}}{4}=\frac{1}{8 \sqrt{7}} \log \left(\frac{\sqrt{7}+(4 x+1)}{\sqrt{7}-(4 x+1)}\right)+c .
\end{aligned}
$
(iii)
$
I=\int \frac{1}{5-6 x-9 x^2} d x
$
$
\begin{aligned}
5-6 x-9 x^2 & =-\left(9 x^2+6 x-5\right) \\
a^2 & =9 x^2 ; \quad a=3 x ; \quad 2 a b=6 x \\
b & =\frac{6 x}{2 a}=\frac{6 x}{2(3 x)}=1 ; \quad b^2=1
\end{aligned}
$
$
\begin{aligned}
& \therefore-\left(9 x^2+6 x-5\right)=-\left(9 x^2+6 x-5+1-1\right) \\
& =-\left\{\left(9 x^2+6 x+1\right)-6\right\}=-\left\{(3 x+1)^2-(\sqrt{6})^2\right\} \\
& =(\sqrt{6})^2-(3 x+1)^2 \\
& \therefore \quad \mathrm{I}=\int \frac{1}{5-6 x-9 x^2} d x=\int \frac{d x}{(\sqrt{6})^2-(3 x+1)^2} \\
& =\frac{\frac{1}{2 \sqrt{6}} \log \frac{\sqrt{6}+(3 x+1)}{\sqrt{6}-(3 x+1)}}{3}=\frac{1}{6 \sqrt{6}} \log \left(\frac{\sqrt{6}+(3 x+1)}{\sqrt{6}-(3 x+1)}\right)+c . \\
&
\end{aligned}
$

Question 2.
(i) $\frac{1}{(2 x+1)^2-16}$
(ii) $\frac{1}{x^2+3 x-3}$
(iii) $\frac{1}{3 x^2-13 x-10}$

Solution:
(i)
$
\begin{aligned}
I=\int \frac{1}{(2 x+1)^2-16} d x & =\int \frac{d x}{(2 x+1)^2-4^2} \\
& =\frac{\frac{1}{2(4)} \log \frac{(2 x+1)-4}{(2 x+1)+4}}{2}=\frac{1}{16} \log \left(\frac{2 x-3}{2 x+5}\right)+c
\end{aligned}
$

(ii)
$
\begin{aligned}
& I=\int \frac{1}{x^2+3 x-3} d x \\
& x^2+3 x-3=x^2+2\left(\frac{3}{2}\right) x-3+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2 \\
& =\left(x+\frac{3}{2}\right)^2-\left(3+\frac{9}{4}\right) \\
& =\left(x+\frac{3}{2}\right)^2-\left(\frac{\sqrt{21}}{2}\right)^2 \\
& \therefore \quad \mathrm{I}=\int \frac{d x}{x^2+3 x-3}=\int \frac{d x}{\left(x+\frac{3}{2}\right)^2-\left(\frac{\sqrt{21}}{2}\right)^2} \\
& =\frac{1}{\frac{2 \sqrt{21}}{2}} \log \left(\frac{x+\frac{3}{2}-\frac{\sqrt{21}}{2}}{x+\frac{3}{2}+\frac{\sqrt{21}}{2}}\right)=\frac{1}{\sqrt{21}}\left(\frac{2 x+3-\sqrt{21}}{2 x+3+\sqrt{21}}\right) \text {. } \\
&
\end{aligned}
$

(iii)
$
\begin{aligned}
& I=\int \frac{1}{3 x^2-13 x-10} d x \\
& 3 x^2-13 x-10=3\left(x^2-\frac{13}{3} x-\frac{10}{3}\right) \\
& =3\left\{x^2-2\left(\frac{13}{6}\right) x-\frac{10}{3}+\left(\frac{13}{6}\right)^2-\left(\frac{13}{6}\right)^2\right\} \\
& =3\left[\left(x-\frac{13}{6}\right)^2-\left(\frac{10}{3}+\frac{169}{36}\right)\right] \\
& =3\left[\left(x-\frac{13}{6}\right)^2-\left(\frac{120+169}{36}\right)\right] \\
& =3\left[\left(x-\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\right] \\
& I=\int \frac{1}{3 x^2-13 x-10} d x=\int \frac{d x}{3\left[\left(x-\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\right]} \\
& =\frac{1}{3}\left\{\frac{1}{2\left(\frac{17}{6}\right)} \log \left(\frac{\left(x-\frac{13}{6}\right)-\frac{17}{6}}{\left.\left(x-\frac{13}{6}\right)+\frac{17}{6}\right)}\right\}\right. \\
& =\frac{1}{17} \log \left(\frac{6 x-30}{6 x+4}\right)=\frac{1}{17} \log \frac{2(3 x-15)}{2(3 x+2)} \\
& =\frac{1}{17} \log \left(\frac{3 x-15}{3 x+2}\right)+c \\
&
\end{aligned}
$

Question 3.
(i) $\frac{1}{\sqrt{(x+1)^2-15}}$
(ii) $\frac{1}{\sqrt{x^2+8 x-20}}$

Solution:
(i) $\int \frac{d x}{\sqrt{(x+1)^2-15}}=\log \left\{(x+1)+\sqrt{(x+1)^2-15}\right\}+c$

(ii)

$
\begin{aligned}
x^2+8 x-20=x^2+8 x-20 & +16-16 \\
& =\left(x^2+8 x+16\right)-36=(x+4)^2-36 \\
\therefore \quad I & =\int \frac{d x}{\sqrt{x^2+8 x-20}}=\log (x+4)+\sqrt{(x+4)^2-36} \\
& =\log \left\{(x+4)+\sqrt{x^2+8 x-20}\right\}+c
\end{aligned}
$