Question

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Evaluate the integral :

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

Solve it.​
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Answer 1

Step-by-step explanation:

First i am direct using formula for integral of arctan since it's well known and substitute limits as t = x and t = 0 , you will get

$\int \: \frac{x \tan {}^{ - 1} (x) - \frac{1}{2} ln(1 + {x}^{2} ) }{ {x}^{3} } dx$

Divide and let saparate our question in two parts ,

first

$\int \: \frac{ \tan {}^{ - 1} (x) }{ {x}^{2} } dx$

integration by parts

$= \frac{ - \tan {}^{ - 1} (x) }{ {x}^{} } + \int \: \frac{dx}{x(1 + {x}^{2} )} \\ \\ = \frac{ - \tan {}^{ - 1} (x) }{ {x}^{} } + \int( \frac{1}{x} - \frac{x}{1 + {x}^{2} } )dx \\ \\ = \frac{ - \tan {}^{ - 1} (x) }{ {x}^{} } + ln(x) - \frac{1}{2} ln(1 + {x}^{2} )$

Now second part

$\frac{ - 1}{2} \int \: \frac{ ln(1 + {x}^{2} ) }{ {x}^{3} } dx \\ \\ put \: \: 1 + {x}^{2} = y \\ xdx = \frac{dy}{2} \\ \\ = \frac{ - 1}{2} \int \: \frac{ ln(y) }{ 2{y}^{2} } dy \\ \\ = \frac{ - 1}{4} \ \ \bigg \{\frac{ ln(y) }{ - y} + \int \: \frac{1}{ {y}^{2} } dy \bigg \} \\ \\ = \frac{1}{4} ( \frac{1 + ln(y) }{y} ) \\ \\ = \frac{1}{4} \frac{1 + ln(1 + {x}^{2} ) }{1 + {x}^{2} }$

For final answer add both parts and put a constant C .