\[\int{\frac{x-1}{{{(x+1)}^{2}}\sqrt{{{x}^{3}}+{{x}^{2}}+x}}dx}\] is equal to

\[3{{\tan }^{-1}}\sqrt{\frac{{{x}^{2}}+x+1}{x}}+C\]

\[{{\tan }^{-1}}\sqrt{\frac{{{x}^{2}}+x+1}{x}}+C\]

\[2{{\tan }^{-1}}\sqrt{\frac{{{x}^{2}}+x+1}{x}}+C\]

\[\int{\frac{x-1}{{{(x+1)}^{2}}\sqrt{{{x}^{3}}+{{x}^{2}}+x}}dx}\] is e

1.	\[\int{\frac{x-1}{{{(x+1)}^{2}}\sqrt{{{x}^{3}}+{{x}^{2}}+x}}dx}\] is equal to
A.	\[{{\tan }^{-1}}\sqrt{\frac{{{x}^{2}}+x+1}{x}}+C\]
B.	\[2{{\tan }^{-1}}\sqrt{\frac{{{x}^{2}}+x+1}{x}}+C\]
C.	\[3{{\tan }^{-1}}\sqrt{\frac{{{x}^{2}}+x+1}{x}}+C\]
D.	None of these
Answer» C. \[3{{\tan }^{-1}}\sqrt{\frac{{{x}^{2}}+x+1}{x}}+C\]

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\[\int{\frac{x-1}{{{(x+1)}^{2}}\sqrt{{{x}^{3}}+{{x}^{2}}+x}}dx}\] is equal to

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