Question

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

• A. $\log \left({\tan }^{-1}\left(x\right)\right)+c$
• B. $\log {\tan }^{-1}{x}^2$
• C. $\log |\sin x+\cos x|+c$
• D. $\log |\sin x-cosx|+c$

Answer 1

The correct option is A $\log \left({\tan }^{-1}\left(x\right)\right)+c$

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

$\left(\frac{1}{t+x^2}\right).dx$ $=d\left({\tan }^{-1}x\right)$

$\int \frac{\left(\frac{1}{1+x^2}\right)}{{\tan }^{-1}x}dx$ $=\int \frac{d\left({\tan }^{-1}x\right)}{{\tan }^{-1}\left(x\right)}$

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

$\int \frac{1}{(1+x^2)\left({\tan }^{-1}x\right)}dx$ $=ln\left({\tan }^{-1}\left(x\right)\right)+c$