Mathematics

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


SOLUTION
$$I=\displaystyle \int \dfrac {\sqrt {a}}{1+x}dx$$
put $$x=t^2$$
$$dx =2t\ dt$$
$$dx=2\sqrt {x}dt$$
$$\dfrac {dx}{\sqrt x}=2dt \Rightarrow \dfrac {\sqrt {x}dx}{x}=2dt$$
$$\Rightarrow \ \sqrt x\ dx=2t^2\ dt$$
$$I=\displaystyle \int \dfrac {2t^2dt}{1+t^2}=\displaystyle \int \dfrac {2(t^2+1)}{1+t^2}dt -2\displaystyle \int \dfrac {dt}{1+t^2}$$
$$=\displaystyle \int 2dt -2\displaystyle \int \dfrac {dt}{1+t^2}$$
$$I=2t-2\tan^{-1}(t)+c$$
$$I=2\sqrt {x}-2\tan^{-1}(\sqrt {x})+c$$

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Subjective Medium Published on 17th 09, 2020
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