Question

If $\int \frac{1}{\left(x+2\right)\left(x^2+1\right)}\mathrm{d}x=a\log \left|1+x^2\right|+b{\tan }^{-1}x+\frac{1}{5}\log \left|x+2\right|+C$, then

$$\begin{gathered} \left(\mathrm{a}\right)a=-\frac{1}{10},b=-\frac{2}{5} \\ \left(\mathrm{b}\right)a=\frac{1}{10},b=-\frac{2}{5} \\ \left(\mathrm{c}\right)a=-\frac{1}{10},b=\frac{2}{5} \\ \left(\mathrm{d}\right)a=\frac{1}{10},b=\frac{2}{5} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{\left(x+2\right)\left(x^2+1\right)}\mathrm{d}x \\ \mathrm{We}\mathrm{express}, \\ \frac{1}{\left(x+2\right)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+1} \\ \Rightarrow 1=A\left(x^2+1\right)+\left(Bx+C\right)\left(x+2\right) \\ \mathrm{On}\mathrm{comparing}\mathrm{the}\mathrm{coefficients}\mathrm{of}{x}^2,x\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 0=A+B\mathrm{and}0=2B+C\mathrm{and}1=A+2C \\ \mathrm{or}A=\frac{1}{5}\mathrm{and}B=-\frac{1}{5}\mathrm{and}C=\frac{2}{5} \\ \therefore I=\int \left(\frac{{\displaystyle \frac{1}{5}}}{x+2}+\frac{-{\displaystyle \frac{1}{5}}x+{\displaystyle \frac{2}{5}}}{x^2+1}\right)\mathrm{d}x \\ =\frac{1}{5}\int \frac{{\displaystyle 1}}{x+2}\mathrm{d}x-\frac{1}{5}\int \frac{{\displaystyle x}}{x^2+1}\mathrm{d}x+\frac{2}{5}\int \frac{{\displaystyle 1}}{x^2+1}\mathrm{d}x \\ =\frac{1}{5}\log \left|x+2\right|-\frac{1}{10}\log \left|x^2+1\right|+\frac{2}{5}{\tan }^{-1}x+C \\ \mathrm{Since},\int \frac{1}{\left(x+2\right)\left(x^2+1\right)}\mathrm{d}x=a\log \left|1+x^2\right|+b{\tan }^{-1}x+\frac{1}{5}\log \left|x+2\right|+C \\ \mathrm{Therefore},a=-\frac{1}{10}\mathrm{and}b=\frac{2}{5}. \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{c}\right). \end{gathered}$$