Exercise 2.7 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $2.7$
Integrate the following with respect to $\mathrm{x}$.
Question 1.
$\frac{1}{9-16 x^2}$
Solution:
$
\frac{1}{9-16 x^2}=\frac{1}{(3)^2-(4 x)^2}
$
Let $4 x=\mathrm{t}, \quad$ Then $4 d x=d t$
So
$
\begin{aligned}
\int \frac{1}{9-16 x^2} d x & =\frac{1}{4} \int \frac{d t}{3^2-t^2} \\
& =\frac{1}{4} \frac{1}{2(3)} \log \left|\frac{3+t}{3-t}\right|+c \\
& =\frac{1}{24} \log \left|\frac{3+4 x}{3-4 x}\right|+c
\end{aligned}
$
Question 2.
$
\frac{1}{9-8 x-x^2}
$
Solution:
$
\int \frac{1}{9-8 x-x^2} d x
$
Consider $9-8 x-x^2=3^2-\left(x^2+8 x\right)$
$
\begin{aligned}
& =9-\left[(x+4)^2-16\right] \\
& =9+16-(x+4)^2 \\
& =25-(x+4)^2 \\
& =5^2-(x+4)^2
\end{aligned}
$
So integral becomes
$
\int \frac{d x}{5^2-(x+4)^2}=\frac{1}{10} \log \left|\frac{5+x+4}{5-x-4}\right|+c
$

$
=\frac{1}{10} \log \left|\frac{9+x}{1-x}\right|+c
$
Question 3.
$
\frac{1}{2 x^2-9}
$
Solution:
$
\int \frac{1}{2 x^2-9} d x=\int \frac{1}{(\sqrt{2 x})^2-3^2} d x
$
Let $\mathrm{t}=\sqrt{2} x$, Then $d t=\sqrt{2} d x$
$
\begin{aligned}
& =\frac{1}{2} \int \frac{d t}{t^2-3^2}=\frac{1}{\sqrt{2}} \frac{1}{2(3)} \log \left|\frac{t-3}{t+3}\right|+c \\
& =\frac{1}{6 \sqrt{2}} \log \left|\frac{\sqrt{2 x}-3}{\sqrt{2 x}+3}\right|+c
\end{aligned}
$

Question 4.
$\frac{1}{x^2-x-2}$
Solution:
$
\int \frac{1}{x^2-x-2} d x
$
Consider $x^2-x-2=x^2-2 x\left(\frac{1}{2}\right)+\frac{1}{4}-\frac{1}{4}-2$
$
\begin{aligned}
& =\left(x-\frac{1}{2}\right)^2-\frac{9}{4} \\
& =\left(x-\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2 \\
\int \frac{d x}{\left(x-\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2} & =\frac{1}{2\left(\frac{3}{2}\right)} \log \left|\frac{x-\frac{1}{2}-\frac{3}{2}}{x-\frac{1}{2}+\frac{3}{2} \mid}\right|+c \\
& =\frac{1}{3} \log \left|\frac{x-2}{x+1}\right|+c
\end{aligned}
$

Question 5.
$\frac{1}{x^2+3 x+2}$
Solution:
$
\begin{aligned}
& \int \frac{1}{x^2+3 x+2} d x \\
& \text { Consider } x^2+3 x+2=\left(x+\frac{3}{2}\right)^2-\frac{9}{4}+2 \\
&=\left(x+\frac{3}{2}\right)^2-\frac{1}{4} \\
&=\left(x+\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2
\end{aligned}
$
So the integral becomes $$ \begin{aligned} \int \frac{d x}{\left(x+\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2} & =\frac{1}{2\left(\frac{1}{2}\right)} \log \left|\frac{x+\frac{3}{2}-\frac{1}{2}}{x+\frac{3}{2}+\frac{1}{2} \mid}\right|+c \\ & =\log \left|\frac{x+1}{x+2}\right|+c\end{aligned} $$
Question 6.
$\frac{1}{2 x^2+6 x-8}$

Solution:
$
\begin{aligned}
\int \frac{1}{2 x^2+6 x-8} & =\frac{1}{2} \int \frac{d x}{x^2+3 x-4} \\
& =\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^2-\frac{9}{4}-4}=\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^2-\frac{25}{4}} \\
& =\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^2-\left(\frac{5}{2}\right)^2} \\
& =\frac{1}{2} \frac{1}{2\left(\frac{5}{2}\right)} \log \left|\frac{x+\frac{3}{2}-\frac{5}{2}}{x+\frac{5}{2}}\right|+c \\
& =\frac{1}{10} \log \left|\frac{x-1}{x+4}\right|+c
\end{aligned}
$

Question 7.
$\frac{e^z}{e^{2 x}-9}$
Solution:
$
\int \frac{e^x d x}{e^{2 x}-9}=\int \frac{e^x d x}{\left(e^x\right)^2-3^2}
$
Let $e^x=t, \quad$ Then $e^x d x=d t$
$
\begin{aligned}
& =\int \frac{d t}{t^2-3^2}=\frac{1}{2(3)} \log \left|\frac{t-3}{t+3}\right|+c \\
& =\frac{1}{6} \log \left|\frac{e^x-3}{e^x+3}\right|+c
\end{aligned}
$
Question 8.
$
\frac{1}{\sqrt{9 x^2-7}}
$
Solution:
$
\begin{aligned}
& \int \frac{1 . d x}{\sqrt{9 x^2-7}}=\int \frac{d x}{\sqrt{9\left(x^2-\frac{7}{9}\right)}}=\frac{1}{3} \int \frac{d x}{\sqrt{x^2-\frac{7}{9}}} \\
& =\frac{1}{3} \int \frac{d x}{\sqrt{x^2-\left(\frac{\sqrt{7}}{3}\right)^2}}=\frac{1}{3} \log \left|x+\sqrt{x^2-\left(\frac{\sqrt{7}}{3}\right)^2}\right|+c
\end{aligned}
$

$
\begin{aligned}
& =\frac{1}{3} \log \left|x+\sqrt{\frac{9 x^2-7}{9}}\right|+c \\
& =\frac{1}{3} \log \left|3 x+\sqrt{9 x^2-7}\right|+c-\frac{1}{3} \log 9 \\
& =\frac{1}{3} \log \left|3 x+\sqrt{9 x^2-7}\right|+k \quad \text { where } k=c-1 / 3 \log 9 \text { which is a constant }
\end{aligned}
$
Question 9.

$
\left(\frac{1}{\sqrt{x^2+6 x+13}}\right)
$
Solution:
$
\begin{aligned}
\int\left(\frac{1}{\sqrt{x^2+6 x+13}}\right) & =\int \frac{d x}{\sqrt{(x+3)^2-9+13}} \\
& =\int \frac{d x}{\sqrt{(x+3)^2+4}} \\
& =\int \frac{d x}{\sqrt{(x+3)^2+2^2}} \\
& =\log \left|(x+3)+\sqrt{(x+3)^2+2^2}\right|+c \\
& =\log \left|x+3+\sqrt{x^2+6 x+13}\right|+c
\end{aligned}
$

Question 10.
$
\left(\frac{1}{\sqrt{x^2-3 x+2}}\right)
$
Solution:
$
\begin{aligned}
\int \frac{d x}{\sqrt{x^2-3 x+2}} & =\int \frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\frac{9}{4}+2}} \\
& =\int \frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}} \\
& =\log \left|\left(x-\frac{3}{2}\right)+\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right|+c \\
& =\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^2-3 x+2}\right|+c
\end{aligned}
$
Question 11.
$
\frac{x^3}{\sqrt{x^8-1}}
$
Solution:
$
\int \frac{x^3}{\sqrt{x^8-1}} d x=\int \frac{x^3 d x}{\sqrt{\left(x^4\right)^2-1^2}}
$
Let $x^4=t, \quad$ Then $4 x^3 d x=d t$
$
\begin{aligned}
& =\frac{1}{4} \int \frac{d t}{\sqrt{t^2-1}}=\frac{1}{4} \log \left|t+\sqrt{t^2-1}\right|+c \\
& =\frac{1}{4} \log \left|x^4+\sqrt{x^8-1}\right|+c
\end{aligned}
$
Question 12.
$\sqrt{1+x+x^2}$

Solution:
$
\begin{aligned}
\int \sqrt{1+x+x^2} d x & =\int \sqrt{\left(x+\frac{1}{2}\right)^2-\frac{1}{4}+1} d x \\
& =\int \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} d x \\
& =\int \sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d x \\
& =\frac{\left(x+\frac{1}{2}\right)}{2} \sqrt{1+x+x^2}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{1+x+x^2}\right|+c
\end{aligned}
$
$\left\{\right.$ Using the formula for $\int \sqrt{x^2+a^2} d x$ with ' $x$ ' as $x+\frac{1}{2}$ and ' $a$ ' as $\left.\frac{\sqrt{3}}{2}\right\}$
Question 13.

$\sqrt{x^2-2}$
Solution:
$
\begin{aligned}
\int \sqrt{x^2-2} d x & =\int \sqrt{x^2-(\sqrt{2})^2 d x} \\
& =\frac{x}{2} \sqrt{x^2-2}-\log \left|x+\sqrt{x^2-2}\right|+c
\end{aligned}
$
Question 14.

$\sqrt{4 x^2-5}$
Solution:

$
\int \sqrt{4 x^2-5} d x=\int \sqrt{(2 x)^2-(\sqrt{5})^2 d x}
$
Let $2 x=t$ Then $2 d x=d t$
$
\begin{aligned}
= & \frac{1}{2} \int \sqrt{t^2-(\sqrt{5})^2 d t} \\
= & \frac{1}{2}\left[\frac{t}{2} \sqrt{t^2-5}-\frac{5}{2} \log \left|t+\sqrt{t^2-5}\right|\right]+c \\
=\frac{1}{4}\left[2 x \sqrt{4 x^2-5}-5\right. & \left.\log \left|2 x+\sqrt{4 x^2-5}\right|\right]+c
\end{aligned}
$
Question 15 .
$\sqrt{2 x^2+4 x+1}$
Solution:
$
\begin{aligned}
\int \sqrt{2 x^2+4 x+1} d x & =\int \sqrt{2\left(x^2+2 x+\frac{1}{2}\right)} d x \\
& =\sqrt{2} \int \sqrt{(x+1)^2-1+\frac{1}{2}} d x \\
& =\sqrt{2} \int \sqrt{(x+1)^2-\frac{1}{2}} d x \\
& =\sqrt{2} \int \sqrt{(x+1)^2-\left(\frac{1}{\sqrt{2}}\right)^2} d x \\
& =\sqrt{2}\left[\frac{(x+1)}{2} \sqrt{(x+1)^2-\frac{1}{2}}-\frac{1}{2} \log \left|(x+1)+\sqrt{(x+1)^2-\frac{1}{2}}\right|\right]+c \\
& =\sqrt{2}\left[\frac{(x+1)}{2} \sqrt{\frac{2 x^2+4 x+1}{\sqrt{2}}}-\frac{1}{4} \log \left|(x+1)+\frac{\sqrt{2 x^2+4 x+1}}{\sqrt{2}}\right|\right]+c \\
& \left.=\left(\frac{x+1}{2}\right)\right] \sqrt{2 x^2+4 x+1}-\frac{\sqrt{2}}{4} \log \left|\sqrt{2}(x+1)+\sqrt{2 x^2+4 x+1}\right|+c
\end{aligned}
$

Question 16.
$
\frac{1}{x+\sqrt{x^2-1}}
$
Solution:

$\int \frac{1 d x}{x+\sqrt{x^2-1}}$
By rationalization,
$
\begin{aligned}
\frac{1}{x+\sqrt{x^2-1}} & =\frac{1}{x+\sqrt{x^2-1}} \times \frac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}} \\
& =\frac{x-\sqrt{x^2-1}}{x^2-\left(x^2-1\right)}=x-\sqrt{x^2-1}
\end{aligned}
$
So the integral becomes,
$
\begin{aligned}
\int x-\sqrt{x^2-1} d x & =\int x d x-\int \sqrt{x^2-1} d x \\
& =\frac{x^2}{2}-\frac{x}{2} \sqrt{x^2-1}+\frac{1}{2} \log \left|x+\sqrt{x^2-1}\right|+c
\end{aligned}
$