Question

$ \int{\frac{4x}{({{x}^{2}}+1)({{x}^{2}}+3)}}dx $ is equal to:

• A. $ \log \left[ \frac{{{x}^{2}}+1}{{{x}^{2}}+3} \right]+c $
• B. $ \log \left[ \frac{{{x}^{2}}+3}{{{x}^{2}}+1} \right]+c $
• C. $ {{\tan }^{-1}}x+\left( \frac{1}{\sqrt{3}} \right){{\tan }^{-1}}\left( \frac{x}{\sqrt{3}} \right)+c $
• D. $ \left( \frac{4}{\sqrt{3}} \right){{\tan }^{-1}}x{{\tan }^{-1}}\left( \frac{x}{\sqrt{3}} \right)+c $
• E. $ 2\log ({{x}^{2}}+1)({{x}^{2}}+3)+c $

Answer 1

Answer: (A)

Let $ I=\int{\frac{4x}{({{x}^{2}}+1)({{x}^{2}}+3)}}dx=\int{\frac{2x}{{{x}^{2}}+1}}dx $ $ -\int{\frac{2x}{{{x}^{2}}+3}}dx $ Let $ {{x}^{2}}+1=u $ and $ {{x}^{2}}+3=v $ $ \Rightarrow $ $ 2x\text{ }dx=du $ and $ 2x\text{ }dx=dv $ $ \therefore $ $ I=\int{\frac{du}{u}}-\int{\frac{du}{v}} $ $ =log\text{ }u-log\text{ }v+c $ $ =\log \frac{({{x}^{2}}+1)}{({{x}^{2}}+3)}+c $