Mathematics

Evaluate $$\int_{0}^{1} x \tan^{-1} x\ dx$$.


SOLUTION
$$\displaystyle \int f(x) g(x) dx= f(x) \displaystyle \int g(x) dx= \displaystyle \int  (f(x) \displaystyle \int  g(x)dx)dx$$
$$\displaystyle \int_0^1 x\tan^{-1}x\ dx=\displaystyle \int_0^1 (\tan^{-1}x)x\ dx$$
$$=\tan^{-1}x \displaystyle \int_0^1x\ dx-\displaystyle \int_0^1\left (\dfrac {d(\tan^{-1}x)}{dx} \displaystyle \int x. dx\right)  dx$$
$$=\left [\tan^{-1}x. \dfrac {x^2}{2}\right]_0^1 -\displaystyle \int_0^1 \dfrac {1}{1+x^2}dx$$
$$=\left [\dfrac {x^2}{2}\tan^{-1}x\right]-\dfrac {1}{2} \displaystyle \int_0^1 \dfrac {x^2 +1-1}{x^2 +1}dx$$
$$=\left [\dfrac {x^2}{2}\tan^{-1}x\right]_0^1 -\dfrac {1}{2} \left [\displaystyle \int_0^1 \dfrac {x^2 +1}{x^2 +1}dx -\displaystyle \int_0^1 \dfrac {dx}{x^2 +1}\right]$$
$$=\left [\dfrac {x^2}{2}\tan^{-1}x\right]-\dfrac {1}{2} \left [\displaystyle \int_0^1 dx-\displaystyle \int_0^1 \dfrac {dx}{x^2 +1}\right]$$
$$\left [Using\ \displaystyle \int \dfrac {dx}{a^2 +x^2}=\dfrac {1}{a} \tan^{-1} \dfrac {x}{a}+C\right]$$
$$=\left [\dfrac {x^2}{2}\tan^{-1}x-\dfrac {1}{2}x +\dfrac {1}{2}+\dfrac {1}{1}\tan^{-1}\dfrac {x}{1}\right]_0^1 C$$
$$=\left [\dfrac {x^2}{2}\tan^{-1}x -\dfrac {x}{2}+\dfrac {1}{2}\tan^{-1}x\right]_0^1$$
$$=\left[ \dfrac { { \left( 1 \right)  }^{ 2 } }{ 2 } \tan ^{ 1 }{ \left( 1 \right)  } -\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \tan ^{ -1 }{ 1 }  \right] -\left [\dfrac {0^2}{2}\tan^{-1}0-\dfrac {0}{2}+\dfrac {1}{2}\tan^{-1} (0)\right]$$
$$=\left [\dfrac {\pi}{8}-\dfrac {1}{2}+\dfrac {\pi}{8}\right]-[0-0+0]$$
$$= \pi /4 -\dfrac {1}{2}$$
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Subjective Medium Published on 17th 09, 2020
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