Question

$\int \frac{(ta{n}^{-1}x{)}^3}{(1+x^2)}dx$ is equal to

• A. $3({\tan }^{-1}x{)}^2+c$
• B. $\frac{(ta{n}^{-1}x{)}^4}{4}+c$
• C. $({\tan }^{-1}x{)}^4+c$
• D. None of these

Answer 1

Answer: (B)

$\frac{(ta{n}^{-1}x{)}^4}{4}+c$

$I=\int \frac{({\tan }^{-1}x{)}^3}{(1+x^2)}dx$

We know that, $\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$

$I=\int ({\tan }^{-1}x{)}^3\left(\frac{d}{dx}{\tan }^{-1}x\right)dx$

$I=\int \frac{d}{dx}\left(\frac{1}{4}\right({\tan }^{-1}x{)}^4)dx$

$\therefore$ $I=\frac{({\tan }^{-1}x{)}^4}{4}+c$