Question

evaluate the indefinite integral.
int x^{3} sqrt{10+x^{4}} d x=square + c
question help: video

Explanation:

Step1: Substitution

Let \(u = 10 + x^{4}\), then \(du=4x^{3}dx\), and \(x^{3}dx=\frac{1}{4}du\).

Step2: Integrate

The integral \(\int x^{3}\sqrt{10 + x^{4}}dx\) becomes \(\frac{1}{4}\int\sqrt{u}du\).
Since \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n=\frac{1}{2}\), we have \(\frac{1}{4}\int u^{\frac{1}{2}}du=\frac{1}{4}\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\).
Simplify \(\frac{1}{4}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{6}u^{\frac{3}{2}}+C\).

Step3: Back - substitution

Substitute \(u = 10 + x^{4}\) back, we get \(\frac{1}{6}(10 + x^{4})^{\frac{3}{2}}+C\).

Answer:

\(\frac{1}{6}(10 + x^{4})^{\frac{3}{2}}\)