Mathematics

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

##### SOLUTION
$I=\displaystyle \int_{0}^{1}\dfrac{2x+3}{5x^2+1}dx=\int_{0}^{1}\dfrac{2x}{5x^2+1}dx+\int_{0}^{1}\dfrac{3}{5x^2+1}dx$
$\Rightarrow I=I_1+I_2$
$I_1=\displaystyle \int_{0}^{1}\dfrac{2x}{5x^2+1}dx$
Put $5x^2+1=t\Rightarrow 5\times 2xdx=dt$ and limits will be
Lower limit $x=0\Rightarrow t=1$ and Upper limit $x=1\Rightarrow t=6$
$I_1=\displaystyle \dfrac{1}{5}\int_{1}^{6}\dfrac{1}{t}dt=\dfrac{1}{5}|\ln|t||_1^{6}=\dfrac{1}{5}|\ln 6-\ln 1|=\dfrac{1}{5}\ln 6$
$I_2=\displaystyle \int_{0}^{1}\dfrac{3}{5x^2+1}dx=\dfrac{3}{5}\int_{0}^{1}\dfrac{1}{x^2+\dfrac{1}{5}}dx$
Integral of $\displaystyle \int_{a}^{b}\dfrac{1}{x^2+a^2}dx=\left | \dfrac{1}{a}\tan^{-1}\dfrac{x}{a}\right |_{a}^{b}$
Here, $a=\dfrac{1}{\sqrt{5}}$
$I_2=\dfrac{3}{5}\left |\sqrt{5}\tan^{-1}(\sqrt{5}x) \right |_{0}^{1}=\dfrac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$
$\therefore I= \dfrac{1}{5}\ln 6+\dfrac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

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Q4 Single Correct Hard
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