Question

evaluate the indefinite integral. (use c for the constant of integration.)
$$ int x ^ { 3 } sqrt { x ^ { 2 } + 10 } d x $$

Explanation:

Step1: Use substitution

Let \(u = x^{2}+10\), then \(du=2xdx\) and \(x^{2}=u - 10\).
The integral \(\int x^{3}\sqrt{x^{2}+10}dx=\frac{1}{2}\int x^{2}\sqrt{x^{2}+10}(2xdx)\).
Substitute \(u\) and \(x^{2}\): \(\frac{1}{2}\int(u - 10)\sqrt{u}du\).

Step2: Expand the integrand

\(\frac{1}{2}\int(u - 10)\sqrt{u}du=\frac{1}{2}\int(u^{\frac{3}{2}}-10u^{\frac{1}{2}})du\).

Step3: Integrate term - by - term

Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have:
\(\frac{1}{2}(\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}-10\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1})+C=\frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}-\frac{20}{3}u^{\frac{3}{2}})+C\).

Step4: Substitute back \(u=x^{2}+10\)

\(\frac{1}{5}(x^{2}+10)^{\frac{5}{2}}-\frac{10}{3}(x^{2}+10)^{\frac{3}{2}}+C\).

Answer:

\(\frac{1}{5}(x^{2}+10)^{\frac{5}{2}}-\frac{10}{3}(x^{2}+10)^{\frac{3}{2}}+C\)