indefinite-integration

Question: $\int_{{}}^{{}}{{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}dx=}$

1) $x{{\tan }^{-1}}x+c$

2) $x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+c$

3) $2x{{\tan }^{-1}}x+\log (1+{{x}^{2}})+c$

4) $2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+c$

Solution:

Explanation: No Explanation

Integration by parts

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