Question

evaluate the indefinite integral.

$$ int x ^ { 3 } sqrt { 1 + x ^ { 4 } } d x $$

Explanation:

Step1: Substitute \( u = 1 + x^{4} \)

Differentiate \( u \) with respect to \( x \): \( du=4x^{3}dx \), so \( x^{3}dx=\frac{1}{4}du \)
The integral \( \int x^{3}\sqrt{1 + x^{4}}dx \) becomes \( \frac{1}{4}\int\sqrt{u}du \)

Step2: Integrate \( \sqrt{u} \)

Recall that \( \int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). For \( n=\frac{1}{2}\), \( \int\sqrt{u}du=\int u^{\frac{1}{2}}du=\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{2}{3}u^{\frac{3}{2}}+C \)
So \( \frac{1}{4}\int\sqrt{u}du=\frac{1}{4}\times\frac{2}{3}u^{\frac{3}{2}}+C=\frac{1}{6}u^{\frac{3}{2}}+C \)

Step3: Substitute back \( u = 1 + x^{4} \)

We get \( \frac{1}{6}(1 + x^{4})^{\frac{3}{2}}+C \)

Answer:

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