Question

If $\int \frac{dx}{\left(x+2\right)\left(x^2+1\right)}=alog\left|1+x^2\right|+bta{n}^{-1}x+\frac{1}{5}log\left|x+2\right|+C$, then

• A. $a=\frac{-1}{10},b=\frac{-2}{5}$
• B. $a=\frac{1}{10},b=\frac{-2}{5}$
• C. $a=\frac{-1}{10},b=\frac{2}{5}$
• D. $a=\frac{1}{10},b=\frac{2}{5}$

Answer 1

Answer: (C)

We have, $I=\int \frac{dx}{\left(x+2\right)\left(x^2+1\right)}$

Let $\frac{1}{\left(x+2\right)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+1}$

$\Rightarrow 1=A\left(x^2+1\right)+Bx\left(x+2\right)+C\left(x+2\right)...\left(i\right)$

Put $x=0$ in $\left(i\right)$, we get $A+2C=1$

Put $x=-2$ in $\left(i\right)$, we get $A=\frac{1}{5}$

$\Rightarrow C=\frac{2}{5}$

Put $x=1$ in $\left(i\right)$, we get

$1=2A+3B+3C$, we get $B=\frac{-1}{5}$

$\Rightarrow \int \frac{dx}{\left(x+2\right)\left(x^2+1\right)}=\frac{1}{5}\int \frac{dx}{x+2}-\frac{1}{5}\int \frac{xdx}{x^2+1}+\frac{2}{5}\int \frac{dx}{x^2+1}$

$=\frac{1}{5}log\left|x+2\right|-\frac{1}{10}log\left|x^2+1\right|+\frac{2}{5}ta{n}^{-1}x+C$

Hence, $a=\frac{-1}{10}$ and $b=\frac{2}{5}$