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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
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B
πm12
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C
π21m
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D
None of these
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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

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