Question

Evaluate the integral:

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

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

For solving this integral, we will first solve the nested integral

Let,

$:\longmapsto \displaystyle I_1 = \int \limits_0^x tan^{-1}tdt$

Solving the integral using standard formulaes,

$:\longmapsto I_1 =\displaystyle \left[\dfrac{1}{1+ t^2}\right]^x_0$

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

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

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

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

Now solving the main integral by substituting I1,

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

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

Splitting to solve the integrand,

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

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

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

where C1 is constant of integration

Substituting 1+x² = u in the remaining integral , so that 2xdx = du

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

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

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

where C= C1 + C2

Hence ,

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

Formulas used:

• $\displaystyle \int tan^{-1}xdx = \dfrac{1}{1+x^2} + C$
• $\displaystyle \int \dfrac{1}{x}dx = log|x| + C$