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Question Number 167775 by peter frank last updated on 24/Mar/22

Commented by infinityaction last updated on 24/Mar/22

$$put{x}^4=t$$

Answered by peter frank last updated on 25/Mar/22

$$\mathrm{u}={\mathrm{x}}^4$$$$\mathrm{du}={4\mathrm{x}}^3\mathrm{dx}$$$$\mathrm{dx}=\frac{\mathrm{du}}{{4\mathrm{x}}^3}$$$$\int \frac{{\mathrm{x}}^3}{1+{\mathrm{u}}^2}.\frac{\mathrm{du}}{{4\mathrm{x}}^3}$$$$\frac{1}{4}\int \frac{\mathrm{du}}{1+{\mathrm{u}}^2}$$$$$$

Commented by mr W last updated on 25/Mar/22

$$\frac{1}{4}\int \frac{\mathrm{du}}{1+{\mathrm{u}}^2}=\frac{{\tan }^{-1}u}{4}+C$$

Commented by peter frank last updated on 25/Mar/22

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

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