Question

1. $int x^{2}sqrt{x^{3}+7}dx$

Explanation:

Step1: Use substitution

Let $u = x^{3}+7$, then $du=3x^{2}dx$, and $x^{2}dx=\frac{1}{3}du$.

Step2: Rewrite the integral

The original integral $\int x^{2}\sqrt{x^{3}+7}dx$ becomes $\frac{1}{3}\int\sqrt{u}du$.

Step3: Integrate $\sqrt{u}$

Since $\sqrt{u}=u^{\frac{1}{2}}$, and $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n
eq - 1$), then $\frac{1}{3}\int u^{\frac{1}{2}}du=\frac{1}{3}\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C$.

Step4: Simplify the result

$\frac{1}{3}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{9}u^{\frac{3}{2}}+C$.

Step5: Substitute back $u$

Substitute $u = x^{3}+7$ back, we get $\frac{2}{9}(x^{3}+7)^{\frac{3}{2}}+C$.

Answer:

$\frac{2}{9}(x^{3}+7)^{\frac{3}{2}}+C$