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
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