Write a value of ( ^ -1 x)^3 1+x^2 dx - Vidyadip

Question

Write a value of $\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$

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Answer

Let $\text{I}=\int\frac{(\tan^{-1}\text{x})^3}{1+\text{x}^2}\text{dx}$
Let $\tan^{-1}\text{x}=\text{t}$
$\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{t}^3}{1}\text{dt}$
$=\frac{\text{t}^4}{4}+\text{C}$
$\text{I}=\frac{(\tan^{-1}\text{x})^4}{4}+\text{C}$

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