Mathematics

Solve $$\displaystyle\int_0^2 {\frac{{5x + 2}}{{{x^2} + 4}}dx} $$


SOLUTION

Consider the given integral.


$$I=\displaystyle\int_{0}^{2}{\dfrac{5x+2}{{{x}^{2}}+4}}dx$$


$$ I=5\displaystyle\int_{0}^{2}{\dfrac{x}{{{x}^{2}}+4}}dx+\displaystyle\int_{0}^{2}{\dfrac{2}{{{x}^{2}}+4}}dx $$


$$ I={{I}_{1}}+{{I}_{2}} $$


 


Now,


$${{I}_{1}}=5\int_{0}^{2}{\dfrac{x}{{{x}^{2}}+4}}dx$$


 


Let $$t={{x}^{2}}+4$$


$$ \dfrac{dt}{dx}=2x+0 $$


$$ \dfrac{dt}{2}=xdx $$


 


Therefore,


$$ I=\dfrac{5}{2}\displaystyle\int_{4}^{8}{\dfrac{1}{t}}dt $$


$$ I=\dfrac{5}{2}\left[ {{\log }_{e}}\left( t \right) \right]_{4}^{8} $$


$$ I=\dfrac{5}{2}\left( {{\log }_{e}}8-{{\log }_{e}}4 \right) $$


$$ I=\dfrac{5}{2}\left( {{\log }_{e}}2 \right) $$


 


Now,


$${{I}_{2}}=\displaystyle\int_{0}^{2}{\dfrac{2}{{{x}^{2}}+{{2}^{2}}}}dx$$


 


We know that


$$\displaystyle\int{\dfrac{dx}{{{a}^{2}}+{{x}^{2}}}}=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)$$


 


Therefore,


$$ {{I}_{2}}=\dfrac{2}{2}\left[ {{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]_{0}^{2} $$


$$ {{I}_{2}}={{\tan }^{-1}}\left( \dfrac{2}{2} \right)-{{\tan }^{-1}}\left( 0 \right) $$


$$ {{I}_{2}}={{\tan }^{-1}}\left( 1 \right)-{{\tan }^{-1}}\left( 0 \right) $$


$$ {{I}_{2}}={{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)-{{\tan }^{-1}}\left( \tan 0 \right) $$


$$ {{I}_{2}}=\dfrac{\pi }{4} $$


 


Therefore,


$$I={{I}_{1}}+{{I}_{2}}$$


$$I=\dfrac{5}{2}\left( {{\log }_{e}}2 \right)+\dfrac{\pi }{4}$$


 


Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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