Integral Calculus Question 208
Question: $ \int{\frac{x-1}{{{(x+1)}^{2}}\sqrt{x^{3}+x^{2}+x}}dx} $ is equal to
Options:
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
Show Answer
Answer:
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{( 1-\frac{1}{x^{2}} )}{( x+\frac{1}{x}+2 )\sqrt{x+\frac{1}{x}+1}}dx=\int{\frac{2zdz}{(z^{2}+1).z}}} $ $ [ Puttingx+\frac{1}{x}+1=z^{2}\Rightarrow ( 1-\frac{1}{x^{2}} )dx=2zdz ] $ $ =2\int{\frac{dz}{1+z^{2}}=2{{\tan }^{-1}}z+C} $ $ =2{{\tan }^{-1}}\sqrt{\frac{x^{2}+x+1}{x}}+C $
BETA
