Question

# If $$\displaystyle I_{m} = \int_{0}^{1}x^m\tan^{-1}x\ dx$$ then $$\left ( m+1 \right ) I_{m}+ \left ( m-1 \right )I_{m-2}$$ is?

A
π4
B
πm12
C
π21m
D
None of these

Solution

## The correct option is C $$\displaystyle \frac{\pi }{2}-\frac{1}{m}$$$$I_{m}=\displaystyle \int_{0}^{1}x^{m}\tan^{-1}xdx$$$$=\left [\displaystyle\dfrac{x^{m+1}\tan^{-1}x }{m+1}\right ]_{0}^{1}-\displaystyle\dfrac{1}{m+1}\displaystyle \int_{0}^{1}\displaystyle\dfrac{x^{m+1}}{1+x^{2}}dx$$$$\Rightarrow I_{m}\left ( m+1 \right )= \displaystyle\dfrac{\pi}{4}-\displaystyle\displaystyle \int_{0}^{1}\dfrac{x^{m+1}}{1+x^{2}}dx \cdots (i)$$replacing $$m\rightarrow m-2 in (i)$$$$\therefore (m-1)I_{m-2}=\dfrac{\pi }{4}-\displaystyle \int_{0}^{1}\dfrac{x^{m-1}}{1+x^{2}}dx \cdots (ii)$$$$\therefore \quad (m+1)I_{ m }+(m-I)I_{ m-2 }$$$$\dfrac { \pi }{ 4 } -\displaystyle \int _{ 0 }^{ 1 } \dfrac { x^{ m+1 } }{ 1+x^{ 2 } } dx+\dfrac { \pi }{ 4 } -\displaystyle \int _{ 0 }^{ 1 } \dfrac { x^{ m-1 } }{ 1+x^{ 2 } } dx$$$$\dfrac { \pi }{ 2 } -\displaystyle \int _{ 0 }^{ 1 } \dfrac { x^{ m+1 }+x^{ m-1 } }{ 1+x^{ 2 } } dx$$$$=\dfrac { \pi }{ 2 } -\displaystyle \int _{ 0 }^{ 1 } \dfrac { x^{ m-1 }\left( 1+x^{ 2 } \right) }{ 1+x^{ 2 } } dx$$$$=\displaystyle\dfrac{\pi }{2}-\displaystyle \int_{0}^{1}x^{m-1}dx$$$$=\displaystyle\dfrac{\pi }{2}-\dfrac{1}{m}$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More