Question

The value of integral $\underset{0}{\overset{3}{\int }}\left({\tan }^{-1}\left(\frac{x}{1+x^2}\right)+{\tan }^{-1}\left(\frac{x^2+1}{x}\right)\right)dx$

• A. $\frac{3\pi }{2}$
• B. $\pi$
• C. $2\pi$
• D. None of these

Answer 1

Answer: (A)

$\frac{3\pi }{2}$

Let,

$y={\int }_0^3{\tan }^{-1}\frac{x}{1+x^2}+{\tan }^{-1}\frac{1+x^2}{x}dx$

$\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$

$\therefore y={\int }_0^3{\tan }^{-1}\left(\frac{\frac{x}{1+x^2}+\frac{1+x^2}{x}}{1-\frac{x}{1+x^2}\times \frac{1+x^2}{x}}\right)dx$

$\therefore y={\int }_0^3{\tan }^{-1}\left(\frac{\frac{x}{1+x^2}+\frac{1+x^2}{x}}{1-1}\right)dx$

$\therefore y={\int }_0^3{\tan }^{-1}\left(\frac{\frac{x}{1+x^2}+\frac{1+x^2}{x}}{0}\right)dx$

$y={\int }_0^3{\tan }^{-1}\infty dx$

$y={\int }_0^3{\tan }^{-1}\tan \frac{\pi }{2}dx$

$y={\int }_0^3\frac{\pi }{2}dx$

$y=\frac{\pi }{2}{\int }_2^{3}1dx=\frac{\pi }{2}{{\left[x\right]}^3}_0$

$y=\frac{\pi }{2}[3-0]$

$y=\frac{\pi }{2}\times 3$

$y=\frac{3\pi }{2}$