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,

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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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