Question

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

• A. $\frac{\pi }{4}$
• B. $2\pi$
• C. $\pi$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (B)

$2\pi$

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

$=I={\int }_{-1}^{3}ta{n}^{-1}\left(\frac{\frac{x}{x^2+1}+\frac{x^2+1}{x}}{1-\frac{x}{x^2+1}.\frac{x^2+1}{x}}\right)dx$ $[\because ta{n}^{-1}x+ta{n}^{-1}y=tan\frac{x+y}{1-xy}]$

$=I={\int }_{-1}^{3}ta{n}^{-1}\left(\frac{\frac{x}{x^2+1}+\frac{x^2+1}{x}}{1-1}\right)dx$

$=I={\int }_{-1}^{3}ta{n}^{-1}\left(\frac{\frac{x}{x^2+1}+\frac{x^2+1}{x}}{0}\right)dx$

$={\int }_{-1}^{3}ta{n}^{-1}\infty dx$

$=I={\int }_{-1}^{3}ta{n}^{-1}tan\frac{\pi }{2}dx$

$=I={\int }_{-1}^3\frac{\pi }{2}dx=\frac{\pi }{2}[x{]}_{-1}^3=\frac{\pi }{2}[3+1]=\frac{\pi }{2}\times 4$

$=2\pi$

$\therefore I=2\pi$