Mathematics

If $$\int \frac { x ^ { 2 } \tan ^ { - 1 } x } { 1 + x ^ { 2 } } d x = \tan ^ { - 1 } x - \frac { 1 } { 2 } \log \left( 1 + x ^ { 2 } \right) + f ( x ) + c$$ then $$f ( x ) =$$


ANSWER

$$- \frac { 1 } { 2 } \left( \tan ^ { - 1 } x \right) ^ { 2 }$$


SOLUTION

$$\\\int\>(\frac{(1+x^2-1)tan^{-1}x}{1+x^2})dx\\=\int\>\left(1-(\frac{1}{1+x^2})\right)tan^{-1}x\>dx\\=\int\>1.tan^{-1}dx-\int\>(\frac{tan^{-1}x}{1+x^2})dx\\Use\>ILATE\>in\>first\>part\>and\>assume\\tan^{-1}x=t\>for\>second\>integration\\=xtan^{-1}x-\int\>(\frac{1}{1+x^2})\times\>xdx-\int\>t\>dt\\=xtan^{-1}x-(\frac{1}{2})log|1+x^2|-(\frac{t^2}{2})+C\\=xtan^{-1}x-(\frac{1}{2})log|1+x^2|-(\frac{(tan^{-1}x)^2}{2})+C\\\therefore\>C=-(\frac{1}{2})(tan^{-1}x)^2$$

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Single Correct Medium Published on 17th 09, 2020
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