indefinite-integration

Question: $\int_{{}}^{{}}{x{{\tan }^{-1}}}xdx=$



1) $\frac{1}{2}({{x}^{2}}+1){{\tan }^{-1}}x-\frac{1}{2}x+c$
2) $\frac{1}{2}({{x}^{2}}-1){{\tan }^{-1}}x-\frac{1}{2}x+c$
3) $\frac{1}{2}({{x}^{2}}+1){{\tan }^{-1}}x+\frac{1}{2}x+c$
4) $\frac{1}{2}({{x}^{2}}+1){{\tan }^{-1}}x-x+c$
Solution: Explanation: No Explanation
Integration by parts

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