Question

$\int \frac{(1+x{)}^2}{x(1+x^2)}dx=$

• A. $2{\tan }^{-1}x+\log x+c$
• B. $\tan x+\log x+c$
• C. $2{\tan }^{-1}x-\log x+c$
• D. $\tan x-\log x+c$

Answer 1

The correct option is A $2{\tan }^{-1}x+\log x+c$

$I=\int \frac{{(1+x)}^2}{x(1+x^2)}.dx$

$\int \frac{(1+x^2)+2x}{x(1+x^2)}.dx$

$=\int \frac{1}{x}.dx+2\int \frac{1}{1+x^2}.dx$

$=\log \left|x\right|+2{\tan }^{-1}\left(x\right)+c$ ($\because \int \frac{1}{x}.dx=\log \left|x\right|+c$ and $\int \frac{1}{1+x^2}.dx={\tan }^{-1}x+c$ and $\int \frac{{(1+x)}^2}{x(1+x^2)}dx$)

$=\log \left|x\right|+2{\tan }^{-1}\left(x\right)+c$