Question

The value of integral $\int {\tan }^{-1}\left(\frac{x^3}{1+x^2}\right)+{\tan }^{-1}\left(\frac{1+x^2}{x^3}\right)dx$ is equal to

• A. $1$
• B. $-\frac{\pi }{2}+c$
• C. $\frac{\pi }{2}+c$
• D. $\left(\frac{\pi }{2}\right)x+c$

Answer 1

Answer: (D)

$\left(\frac{\pi }{2}\right)x+c$

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

$=\int {\tan }^{-1}\left[\frac{\frac{x^3}{1+x^2}+\frac{1+x^2}{x^3}}{1-\left(\frac{x^3}{1+x^2}\right)\left(\frac{1+x^2}{x^3}\right)}\right]dx$

$=\int {\tan }^{-1}\left(\frac{\frac{x^6+{\left(1+x^2\right)}^2}{\left(1+x^2\right)\left(x^3\right)}}{0}\right)dx$

$=\int \frac{\pi }{2}dx$

$=\frac{\pi }{2}x+c$