Question

If $I_m={\int }_0^1{x}^m{\tan }^{-1}xdx$

then $(m+1)I_m+(m-1)I_{m-2}$ is?

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{m}-\frac{1}{2}$
• C. $\frac{\pi }{2}-\frac{1}{m}$
• D. None of these

Answer 1

The correct option is C $\frac{\pi }{2}-\frac{1}{m}$

$I_m={\int }_0^1{x}^m{\tan }^{-1}xdx$

$={\left[\frac{x^{m+1}{\tan }^{-1}x}{m+1}\right]}_0^1-\frac{1}{m+1}{\int }_0^1\frac{x^{m+1}}{1+x^2}dx$

$\Rightarrow I_m(m+1)=\frac{\pi }{4}-{\int }_0^1\frac{x^{m+1}}{1+x^2}dx\cdots \left(i\right)$

replacing $m\to m-2in\left(i\right)$

$\therefore (m-1)I_{m-2}=\frac{\pi }{4}-{\int }_0^1\frac{x^{m-1}}{1+x^2}dx\cdots \left(ii\right)$

$\therefore (m+1)I_m+(m-I)I_{m-2}$

$\frac{\pi }{4}-{\int }_0^1\frac{x^{m+1}}{1+x^2}dx+\frac{\pi }{4}-{\int }_0^1\frac{x^{m-1}}{1+x^2}dx$

$\frac{\pi }{2}-{\int }_0^1\frac{x^{m+1}+x^{m-1}}{1+x^2}dx$

$=\frac{\pi }{2}-{\int }_0^1\frac{x^{m-1}(1+x^2)}{1+x^2}dx$

$=\frac{\pi }{2}-{\int }_0^1{x}^{m-1}dx$

$=\frac{\pi }{2}-\frac{1}{m}$