Question

If $ \int\frac{dx}{\left(x+2\right)\left(x^{2} +1\right)} = a log\left|1+x^{2}\right| +b tan^{-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

The Correct Option is 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) \quad...\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 + 3 C$, 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 }tan^{-1} x + C $ Hence, $a= \frac{-1}{10}$ and $b = \frac{2}{5}$