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Question Number 101747 by Dwaipayan Shikari last updated on 04/Jul/20

$$\int \frac{x^{\frac{-1}{2}}}{1+x^{\frac{1}{3}}}dx$$

Answered by bemath last updated on 04/Jul/20

$$setx=t^6\Rightarrow \mathrm{dx}={6\mathrm{t}}^5\mathrm{dt}$$$$\int \frac{{\mathrm{t}}^{-3}}{1+{\mathrm{t}}^2}.{6\mathrm{t}}^5\mathrm{dt}=\int \frac{{6\mathrm{t}}^2}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$\mathrm{set}\mathrm{t}=\tan \theta$$$$\int \frac{6\tan {}^2\theta \sec {}^2\theta \mathrm{d}\theta }{\sec {}^2\theta }=6\tan \theta -{6\tan }^{-1}\left(\theta \right)+\mathrm{C}$$$$=6\mathrm{t}-{6\tan }^{-1}\left(\mathrm{t}\right)+\mathrm{c}$$$$=6\sqrt[6]{x}-6\tan \left(\sqrt[6]{x}\right)+c★$$

Commented by Dwaipayan Shikari last updated on 04/Jul/20

$$\mathit{G}\mathit{r}\mathit{e}\mathit{a}\mathit{t}!$$

Commented by bemath last updated on 04/Jul/20

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

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