Question

$\bf \: Evaluate \: the \: integral: \\ \sf{ \displaystyle \int \frac{ \int \limits_ 0^x \tan^{ -1}t \: dt}{ {x}^{3} } dx}$
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Try to Solve it !!
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Answer 1

${ \bold \red{\displaystyle \int \frac{ \int \limits_0^x \tan^{ -1}t \: dt}{ {x}^{3} } dx}}$

$\bold \red{ \displaystyle I_1 = \int \limits_0^x tan^{-1}tdt}$

$\red{ I_1 =\displaystyle \left[\dfrac{1}{1+ t^2}\right]^x_{0}}$

$\red{ I_1 =\displaystyle \left[\dfrac{1}{1+x^2} - \dfrac{1}{1+0^2}\right]}$

$\red{ I_1 =\dfrac{1}{1+x^2} - 1}$

$\red{I_1=\dfrac{1-1-x^2}{1+x^2}}$

$\red{ I_1=\dfrac{-x^2}{1+x^2}}$

$\red{ I = \displaystyle \int \dfrac{-x^2}{(1+x^2)x^3}}$

$\red{I = -\displaystyle \int \dfrac{1}{x(1+x^2)}}$

$\red{ I =- \displaystyle \int \dfrac{1}{x} - \dfrac{x}{1+x^2} dx}$

$\red{I = -\displaystyle \int \dfrac{1}{x} dx + \int \dfrac{x}{1+x^2}dx}$

$\red{I = \displaystyle - log|x| + C_1 + \int \dfrac{x}{1+x^2}dx}$

$\red{ I = \displaystyle -log|x| + C_1 + \int \dfrac{du}{2u}}$

$\red{ I = -log|x| + C_1 + \dfrac{log|u|}{2} + C_2}$

$\red{ I = -log|x| + \dfrac{log|1+x^2|}{2} + C}$

$\red{{\displaystyle I = log|x| + \dfrac{log|1+x^2|}{2} + C}}$