Mathematics

# Evaluate the definite integral :$\displaystyle \int_{0}^{1} \dfrac {2x+3}{5x^2 +1}dx$

##### SOLUTION
$I=\displaystyle \int_0^1 \dfrac {2x+3}{5x^2 +1}dx =\displaystyle \int_0^1 \dfrac {2x}{5x^2 +1}dx +\displaystyle \int_0^1 \dfrac {3}{5x^2 +1}dx =\dfrac {1}{5} \displaystyle \int_0^1 \dfrac {10x}{5x^2 +1}dx +3\displaystyle \int_0^1 \dfrac {1}{(\sqrt {5} x)^2 +1} dx$
$\Rightarrow \ I=\dfrac {1}{5}[\log (5x^2 +1)]_0^1 +\dfrac {3}{\sqrt 5}\left [\tan^{-1} \dfrac {\sqrt 5 x}{1}\right]_0^1$
$\Rightarrow \ I=\dfrac {1}{5}(\log 6-\log 1)+\dfrac {3}{\sqrt 5}\tan^{-1}\sqrt 5 =\dfrac {1}{5}\log 6+\dfrac {3}{\sqrt 5}\tan^{-1}\sqrt 5$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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