Exercise 2.6 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $2.6$
Integrate the following with respect to $\mathrm{x}$.
Question 1.
$\frac{2 x+5}{x^2+5 x-7}$
Solution:
$
\frac{2 x+5}{x^2+5 x-7}
$
Let $f(x)=x^2+5 x-7 \quad$ Then $f^{\prime}(x)=2 x+5$
So $\int \frac{2 x+5}{x^2+5 x-7} d x=\int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c$
$
=\log \left|x^2+5 x-7\right|+c
$
Question 2.
$\frac{e^{3 \log z}}{x^4+1}$
Solution:
$
\frac{e^{3 \log x}}{x^4+1}=\frac{e^{\log x^3}}{x^4+1}=\frac{x^3}{x^4+1}
$
Let $x^4+1=f(x) \quad$ Then $4 x^3=f^{\prime}(x)$
So $\int \frac{e^{3 \log x}}{x^4+1} d x=\int \frac{x^3}{x^4+1} d x=\frac{1}{4} \int \frac{4 x^3}{x^4+1} d x$
$
\begin{aligned}
& =\frac{1}{4} \int \frac{f^{\prime}(x)}{f(x)} d x=\frac{1}{4} \log |f(x)|+c \\
& =\frac{1}{4} \log \left|x^4+1\right|+c
\end{aligned}
$
Question 3.
$
\frac{e^{2 x}}{e^{2 x}-2}
$
Solution:

$\frac{e^{2 x}}{e^{2 x}-2}$
Let $f(x)=e^{2 x}-2, \quad$ Then $f^{\prime}(x)=2 e^{2 x}$
So
$
\begin{aligned}
\int \frac{e^{2 x}}{e^{2 x}-2} d x & =\frac{1}{2} \int \frac{2 e^{2 x}}{e^{2 x}-2} d x \\
& =\frac{1}{2} \int \frac{f^{\prime}(x)}{f(x)} d x=\frac{1}{2} \log |f(x)|+c \\
& =\frac{1}{2} \log \left|e^{2 x}-2\right|+c
\end{aligned}
$
Question 4.
$
\frac{(\log x)^3}{x}
$
Solution:
$
\frac{(\log x)^3}{x}
$
Let $\log x=Z, \quad$ Then $\frac{1}{x} d x=d z$
So $\int \frac{(\log x)^3}{x} d x=\int z^3 d z=\frac{z^4}{4}+c$
$
=\frac{(\log x)^4}{4}+c
$
Question 5.
$\frac{6 x+7}{\sqrt{3 x^2+7 x-1}}$
Solution:
$\frac{6 x+7}{\sqrt{3 x^2+7 x-1}}$
Let $f(x)=3 x^2+7 x-1, \quad$ Then $f^{\prime}(x)=6 x+7$
So $\int \frac{6 x+7}{\sqrt{3 x^2+7 x-1}} d x=\int \frac{f^{\prime}(x)}{\sqrt{f(x)}} d x$

$
\begin{aligned}
& =2 \sqrt{f(x)+c} \\
& =2 \sqrt{3 x^2+7 x-1}+c
\end{aligned}
$
Question 6.
$
(4 x+2) \sqrt{x^2+x+1}
$
Solution:
$
(4 x+2) \sqrt{x^2+x+1}
$
Let $f(x)=x^2+x+1$
then $f^{\prime}(x)=2 x+1$
$
\text { So } \begin{aligned}
\int(4 x+2) \sqrt{x^2+x+1} d x & =2 \int(2 x+1) \sqrt{x^2+x+1} d x \\
& =2 \int f^{\prime}(x) \sqrt{f(x)} d x \\
& =2 \int[f(x)]^{\frac{1}{2}} f^{\prime}(x) d x \\
& =2 \frac{[f(x)]^{\frac{3}{2}}}{\frac{3}{2}}+c \\
& =\frac{4}{3}\left(x^2+x+1\right)^{\frac{3}{2}}+c
\end{aligned}
$

Question 7.
$x^8\left(1+x^9\right)^5$
Solution:
$
x^8\left(1+x^9\right)^5
$
Let $f(x)=1+x^9, \quad$ Then $f^{\prime}(x)=9 x^8$
So $\int x^8\left(1+x^9\right)^5 d x=\frac{1}{9} \int 9 x^8\left(1+x^9\right)^5 d x$
$
\begin{aligned}
& =\frac{1}{9} \int[f(x)]^5 f^{\prime}(x) d x \\
& =\frac{1}{9} \frac{[f(x)]^6}{6}+c \\
& =\frac{1}{54}\left(1+x^9\right)^6+c
\end{aligned}
$
Question 8.
$\frac{x^{e-1}+e^{x-1}}{x^e+e^x}$

Solution:
$
\begin{aligned}
\frac{x^{e-1}+e^{x-1}}{x^e+e^x} & =\frac{x^{e-1}+\frac{e^x}{e}}{x^e+e^x} \\
& =\frac{e x^{e-1}+e^x}{e\left(x^e+e^x\right)}
\end{aligned}
$
Let $f(x)=x^e+e^x \quad$ Then $f^{\prime}(x)=e x^{e-1}+e^x$
So
$
\begin{aligned}
\int \frac{x^{e-1} e^{x-1}}{x^e e^x} d x & =\int \frac{e x^{e-1}+e^x}{e\left(x^e+e^x\right)} d x \\
& =\frac{1}{e} \int \frac{f^{\prime}(x)}{f(x)} d x \\
& =\frac{1}{e} \log |f(x)|+c=\frac{1}{e} \log \left|x^e+e^x\right|+c
\end{aligned}
$
Question 9.
$\frac{1}{x \log x}$
Solution:
$
\frac{1}{x \log x}
$
Let $\log x=z \quad$ Then $\frac{1}{x} d x=d z$
So $\int \frac{1}{x \log x} d x=\int \frac{d z}{z}=\log z+c$
$
=\log |\log x|+c
$
Question 10.
$
\frac{x}{2 x^4-3 x^2-2}
$
Solution:

$
\frac{x}{2 x^4-3 x^2-2}=\frac{x}{\left(2 x^2+1\right)\left(x^2-2\right)}
$
Let $x^2=t, \quad$ Then $2 x d x=d t$
So $\quad \int \frac{x}{2 x^4-3 x^2-2}=\frac{1}{2} \int \frac{2 x d x}{\left(2 x^2+1\right)\left(x^2-2\right)}$
$
=\frac{1}{2} \int \frac{d t}{(2 t+1)(t-2)}
$
We make use of the partial fraction method
Let $\frac{1}{(2 t+1)(t-2)}=\frac{\mathrm{A}}{2 t+1}+\frac{\mathrm{B}}{t-2}$
$
1=\mathrm{A}(t-2)+\mathrm{B}(2 t+1)
$
Let $t=2$, then $1=5 \mathrm{~B} \Rightarrow \mathrm{B}=\frac{1}{5}$
Let $t=-\frac{1}{2}$, then $1=\frac{-5}{2} \mathrm{~A} \Rightarrow \mathrm{A}=\frac{-2}{5}$
So $\frac{1}{2} \int \frac{d t}{(2 t+1)(t-2)}=\frac{1}{2} \int \frac{\frac{-2}{5}}{2 t+1} d t+\frac{1}{2} \int \frac{\frac{1}{5}}{t-2} d t$
$
=\frac{-\log |2 t+1|}{5(2)}+\frac{1}{10} \log |t-2|+c
$
putting $t=x^2$, we get $=\frac{-\log \left|2 x^2+1\right|}{10}+\frac{1}{10} \log \left|x^2-2\right|+c$
$
=\frac{1}{10} \log \left|\frac{x^2-2}{2 x^2+1}\right|+c \quad \text { Using } \log (a)-\log (b)=\log \frac{a}{b}
$

Question 11.
$\mathrm{e}^{\mathrm{x}}(1+\mathrm{x}) \log \left(\mathrm{x} \mathrm{e}^{\mathrm{x}}\right)$
Solution:
$
e^x(1+x) \log \left(x e^x\right)=\left(e^x+x e^x\right) \log \left(x e^x\right)
$
Let $\mathrm{z}=\mathrm{x} \mathrm{e}^{\mathrm{x}}$, Then $\mathrm{dz}=\mathrm{d}\left(\mathrm{x} \mathrm{e}^{\mathrm{x}}\right)$
$\mathrm{dz}=\left(\mathrm{x} \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}$ (Using product rule)
So $\int e^x(1+x) \log \left(x e^x\right) d x$
$
\begin{aligned}
& =\int \log \left(x e^x\right)\left(e^x+x e^x\right) d x \\
& =\int \log z d z \\
& =z(\log z-1)+c \\
& =x e^x\left[\log \left(x e^x\right)-1\right]+c
\end{aligned}
$

Question 12.
$\frac{1}{x^2\left(x^2+1\right)}$
Solution:
$
\frac{1}{x^2\left(x^2+1\right)}=\frac{x}{x^2\left(x^2+1\right)}
$
Let $x^2=t$, Then $2 x d x=d t$
$
\begin{aligned}
\text { So } \int \frac{1}{x^2\left(x^2+1\right)} d x & =\frac{1}{2} \int \frac{2 x d x}{x^2\left(x^2+1\right)}=\frac{1}{2} \int \frac{d t}{t(t+1)} \\
& =\frac{1}{2} \int\left[\frac{1}{t}-\frac{1}{t+1}\right] d t \text { (Using partial fractions) } \\
& =\frac{1}{2}[\log |t|-\log |t+1|]+c \\
& =\frac{1}{2} \log \left|\frac{t}{t+1}\right|
\end{aligned}
$
Replacing $t$ by $x^2$ we get $=\frac{1}{2} \log \left|\frac{x^2}{x^2+1}\right|+c$
$
=\log |x|-\frac{1}{2} \log \left|x^2+1\right|+c
$
Question 13.
$
e^x\left[\frac{1}{x^2}-\frac{2}{x^3}\right]
$

Solution:
$e^x\left[\frac{1}{x^2}-\frac{2}{x^3}\right] \quad$ Let $f(x)=\frac{-1}{x^2}$ Then $f^{\prime}(x)=\frac{-2}{x^3}$
So $\int e^x\left[\frac{1}{x^2}-\frac{2}{x^3}\right] d x=\int e^x\left[f(x)+f^{\prime}(x)\right] d x$
$
\begin{aligned}
& =e^x f(x)+c \\
& =\frac{e^x}{x^2}+c
\end{aligned}
$
Question 14.
$
e^x\left[\frac{x-1}{(x+1)^3}\right]
$
Solution:
$
\begin{aligned}
e^x\left[\frac{x-1}{(x+1)^3}\right] & =e^x\left[\frac{x+1-1-1}{(x+1)^3}\right] \\
& =e^x\left[\frac{x+1}{(x+1)^3}-\frac{2}{(x+1)^3}\right] \\
& =e^x\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}\right]
\end{aligned}
$
Let $f(x)=\frac{1}{(x+1)^2}, \quad f^{\prime}(x)=\frac{-2}{(x+1)^3}$
So $\begin{aligned} \int \frac{e^x(x-1)}{(x+1)^3} d x & =\int e^x\left[f(x)+f^{\prime}(x)\right] d x \\ & =e^x f(x)+c\end{aligned}$
$
\begin{aligned}
& =e^x f(x)+c \\
& =\frac{e^x}{(x+1)^2}+c
\end{aligned}
$

Question 15.
$
e^{3 x}\left[\frac{3 x-1}{9 x^2}\right]
$
Solution:
$
\begin{aligned}
e^{3 x}\left[\frac{3 x-1}{9 x^2}\right] & =e^{3 x}\left[\frac{1}{3 x}-\frac{1}{9 x^2}\right] \\
& =\frac{e^{3 x}}{9}\left[3\left(\frac{1}{x}\right)+\frac{-1}{x^2}\right]
\end{aligned}
$
Let $f(x)=\frac{1}{x} \quad \therefore f^{\prime}(x)=\frac{-1}{x^2}$
$
\begin{aligned}
\operatorname{So} \int e^{3 x}\left[\frac{3 x-1}{9 x^2}\right] d x & =\frac{1}{9} \int e^{3 x}\left[3\left(\frac{1}{x}\right)-\frac{1}{x^2}\right] d x \\
& =\frac{1}{9} \int e^{a x}\left[a f(x)+f^{\prime}(x)\right] d x \\
& =\frac{1}{9}\left[e^{a x} f(x)\right]+c \\
& =\frac{1}{9}\left[e^{3 x}\left(\frac{1}{x}\right)\right]+c \\
& =\frac{e^{3 x}}{9 x}+c
\end{aligned}
$