Question

b) Obtain \int tan ^{-1\left ( \frac{2x}{1-x^2} \right )}dx.

11 Mar 2024
Answer :
Word Count : 226

To evaluate the integral:

\[ \int \tan^{-1}\left(\frac{2x}{1-x^2}\right) \, dx \]

we can use the substitution method.

Let \( u = \frac{2x}{1-x^2} \).

Then,

\[ du = \frac{d}{dx} \left(\frac{2x}{1-x^2}\right) \, dx = \frac{2(1-x^2) - 2x(-2x)}{(1-x^2)^2} \, dx = ______ _______ _______ _________ ________ _________ ___ _______ ________ _____ ___.
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