Question

The value of $ \int{\frac{{{x}^{2}}+1}{{{x}^{4}}-{{x}^{2}}+1}}dx $ is

• A. $ {{\tan }^{-1}}(2{{x}^{2}}-1)+c $
• B. $ {{\tan }^{-1}}\frac{{{x}^{2}}+1}{x}+c $
• C. $ {{\sin }^{-1}}\left( x-\frac{1}{x} \right)+c $
• D. $ {{\tan }^{-1}}{{x}^{2}}+c $
• E. $ {{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{x} \right)+c $

Answer 1

Answer: (E)

Let $ I=\int{\frac{{{x}^{2}}+1}{{{x}^{4}}-{{x}^{2}}+1}}dx $
$=\int{\frac{1+\frac{1}{{{x}^{2}}}}{{{x}^{2}}+\frac{1}{{{x}^{2}}}-1}}dx $
$=\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)}{{{\left( x-\frac{1}{x} \right)}^{2}}+1}}dx $ Let $ x-\frac{1}{x}=t $ and $ \left( 1+\frac{1}{{{x}^{2}}} \right)dx=dt $
$ \therefore $ $ I=\int{\frac{dt}{{{t}^{2}}+1}}={{\tan }^{-1}}t+c $
$={{\tan }^{-1}}\left( x-\frac{1}{x} \right)+c $
$={{\tan }^{-1}}\left( \frac{{{x}^{2}}-1}{x} \right)+c $