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Question
Write a value of\[\int\frac{\left( \tan^{- 1} x \right)^3}{1 + x^2} dx\]
Solution
\[\text{ Let I} = \int\frac{\left( \tan^{- 1} x \right)^3}{1 + x^2}dx\]
\[\text{ Let tan}^{- 1} x = t\]
\[ \Rightarrow \frac{1}{1 + x^2}dx = dt\]
\[ \therefore I = \int t^3 . dt\]
\[ = \frac{t^4}{4} + C\]
\[ = \frac{\left( \tan^{- 1} x \right)^4}{4} + C \left( \because t = \tan^{- 1} x \right)\]
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