Mathematics

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

##### SOLUTION
$\displaystyle \int \dfrac{1}{1+x^2}dx-\displaystyle\int \dfrac{x}{1+x^2}dx$

$\tan^{-1}x-\dfrac{1}{2}\displaystyle\int \dfrac{2x}{1+x^2}dx\cdots(1)$

Consider $\displaystyle \int \dfrac{2x}{1+x^2}dx$

put $x^2=t\implies 2x dx=dt$

$\implies \displaystyle \int \dfrac{1}{1+t}dt\\\log({1+t})\\\log(1+x^2)$

Put it in $(1)$
$\implies \tan^{-1}x-\dfrac{1}{2}\log(1+x^2)+c$

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

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