druczekq4

2021-11-10

Find the indefinite integral and check the result by differentiation.
$\int \frac{{x}^{3}}{\sqrt{1+{x}^{4}}}dx$

Provere

Step 1
Consider the following integral:
$\int \frac{{x}^{3}}{\sqrt{1+{x}^{4}}}dx$
Substitute $1+{x}^{4}=u\to 4{x}^{3}dx=du\to {x}^{3}dx=\frac{1}{4}du$ in the above integral:
$\int \frac{{x}^{3}}{\sqrt{1+{x}^{4}}}dx=\int \frac{1}{\sqrt{u}}×\frac{1}{4}du$
$=\frac{1}{4}\int \frac{1}{\sqrt{u}}du$
$=\frac{1}{4}×\frac{{u}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C$
$=\frac{1}{4}×2\sqrt{u}+C$
$=\frac{\sqrt{u}}{2}+C$
Step 2
Substitute $u=1+{x}^{4}$ in the above equation:
$\int \frac{{x}^{3}}{\sqrt{1+{x}^{4}}}dx=\frac{\sqrt{1+{x}^{4}}}{2}+C$
Differentiation part:
$\frac{d}{dx}\left(\frac{\sqrt{1+{x}^{4}}}{2}+C\right)=\frac{1}{2}×\frac{1}{2\sqrt{1+{x}^{4}}}×\frac{d}{dx}\left(1+{x}^{4}\right)$
$=\frac{1}{4\sqrt{1+{x}^{4}}}×4{x}^{3}$
$=\frac{{x}^{3}}{\sqrt{1+{x}^{4}}}$

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