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
Correct Answer: B
Solution :
[b] \[\int{\frac{x-1}{{{(x+1)}^{2}}\sqrt{{{x}^{3}}+{{x}^{2}}+x}}dx}\] \[=\int{\frac{({{x}^{2}}-1)}{({{x}^{2}}+2x+1).x\sqrt{x+\frac{1}{x}+1}}dx}\] \[=\int{\frac{\left( 1-\frac{1}{{{x}^{2}}} \right)}{\left( x+\frac{1}{x}+2 \right)\sqrt{x+\frac{1}{x}+1}}dx=\int{\frac{2zdz}{({{z}^{2}}+1).z}}}\] \[\left[ Putting\,\,x+\frac{1}{x}+1={{z}^{2}}\Rightarrow \left( 1-\frac{1}{{{x}^{2}}} \right)dx=2zdz \right]\] \[=2\int{\frac{dz}{1+{{z}^{2}}}=2{{\tan }^{-1}}z+C}\] \[=2{{\tan }^{-1}}\sqrt{\frac{{{x}^{2}}+x+1}{x}}+C\]
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