Mathematics

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

##### SOLUTION
$I=\displaystyle \int_{-1}^1\dfrac {1}{x^2+2x+5}dx$

$=\displaystyle \int_{-1}^1\dfrac {1}{x^2+2x+4+1}dx$

$=\displaystyle \int_{-1}^1\dfrac {1}{(x+1)^2 +2^2}dx$

we know that $=\displaystyle \int_{-1}^1\dfrac {1}{(x)^2 +a^2}dx=\dfrac {1}{2}\left [\tan^{-1} \dfrac {x}{a}\right]_{1-}^1$

$=\dfrac {1}{a}\left [\tan^{-1} \dfrac {x+1}{2}\right]_{1-}^1$

$=\dfrac {1}{2}(\tan^{-1} 1-\tan^{-1}0)=\dfrac{1}{2}\times \dfrac{\pi}{4}$

$=\dfrac {\pi}{8}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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