ii. $int xsqrt{2x^{2}+1}dx$

Explanation:

Step1: Use substitution

Let $u = 2x^{2}+1$, then $du=4xdx$, and $xdx=\frac{1}{4}du$.

Step2: Rewrite the integral

The integral $\int x\sqrt{2x^{2}+1}dx$ becomes $\frac{1}{4}\int\sqrt{u}du$. Since $\sqrt{u}=u^{\frac{1}{2}}$, the integral is $\frac{1}{4}\int u^{\frac{1}{2}}du$.

Step3: Integrate using power - rule

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\frac{1}{4}\int u^{\frac{1}{2}}du$, we have $\frac{1}{4}\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{1}{4}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{6}u^{\frac{3}{2}}+C$.

Step4: Substitute back $u$

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

Answer:

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