Mathematics

Solve:
$$\int\frac{1}{{1 + {x^2}}}$$ dx


SOLUTION
$$I=\displaystyle \int \dfrac {dx}{1+x^2}$$
put 
$$x=\tan p$$
$$dx=\sec^2p \ dp$$
$$dx=(1+\tan^2 p)dp$$
$$dx=(1+x^2)dp$$
$$\Rightarrow \ \dfrac {dx}{(1+x^2)}=dp$$
$$I=\displaystyle \int dp\quad =p+c=\tan^{-1}x+c$$
$$\therefore \ \displaystyle \int \dfrac {dx}{1+x^2}=\tan^{-1}x+c$$

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Subjective Medium Published on 17th 09, 2020
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