Mathematics

# Solve$\displaystyle \int_0^1 \dfrac{x^2-2}{x^2+1}dx$

##### SOLUTION
$I=\displaystyle\int_0^1\dfrac{x^2-2}{x^2+1}dx=\int_0^1\dfrac{x^2+1-3}{x^2+1}dx$

$I=\displaystyle\int_0^1\dfrac{x^2+1}{x^2+1}dx-\int_0^1\dfrac{3}{x^2+1}dx$

$I=\displaystyle\int_0^1dx-3\int_0^1\dfrac{1}{x^2+1}dx$

$I=[x]_0^1-3\times [\tan^{-1}x]_0^1$

$I=1-3[\tan^{-1} 1-\tan^{-1}0]$

$I=1-3[\dfrac{\pi}{4}-0]$

$I=1-\dfrac{3\pi}{4}$

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