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Question Number 116590 by bemath last updated on 05/Oct/20

$$\int \frac{\sqrt{\mathrm{x}}}{1+{\mathrm{x}}^3}\mathrm{dx}=?$$

Answered by Olaf last updated on 05/Oct/20

$$u=x^{3/2},du=\frac{3}{2}\sqrt{x}dx=\frac{3}{2}{u}^{1/3}dx$$$$\int \frac{u^{1/3}}{1+u^2}.\frac{2}{3}.\frac{du}{u^{1/3}}=\frac{2}{3}\arctan u=\frac{2}{3}\arctan \left(x^{3/2}\right)$$

Commented by bemath last updated on 05/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by Dwaipayan Shikari last updated on 05/Oct/20

$$\int \frac{\sqrt{\mathrm{x}}}{1+{\mathrm{x}}^3}\mathrm{dx}=\int \frac{{2\mathrm{t}}^2\mathrm{dt}}{1+{\left({\mathrm{t}}^3\right)}^2}$$$$=\frac{2}{3}\int \frac{\mathrm{du}}{1+{\mathrm{u}}^2}{\mathrm{t}}^3=\mathrm{u}$$$$=\frac{2}{3}{\tan }^{-1}\mathrm{u}+\mathrm{C}=\frac{2}{3}{\tan }^{-1}{\mathrm{x}}^{\frac{3}{2}}+\mathrm{C}$$

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