10. $int x^{3}sqrt{4x^{2}-1}mathrm{d}x$

Explanation:

Step1: Use substitution

Let $u = 4x^{2}-1$, then $du=8xdx$ and $x^{2}=\frac{u + 1}{4}$. Also, $x^{3}dx=\frac{1}{8}(u + 1)du$.

Step2: Rewrite the integral

The original integral $\int x^{3}\sqrt{4x^{2}-1}dx=\frac{1}{8}\int(u + 1)\sqrt{u}du=\frac{1}{8}\int(u^{\frac{3}{2}}+u^{\frac{1}{2}})du$.

Step3: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\frac{1}{8}(\frac{2}{5}u^{\frac{5}{2}}+\frac{2}{3}u^{\frac{3}{2}})+C$.

Step4: Substitute back $u = 4x^{2}-1$

$\frac{1}{20}(4x^{2}-1)^{\frac{5}{2}}+\frac{1}{12}(4x^{2}-1)^{\frac{3}{2}}+C$.

Answer:

$\frac{1}{20}(4x^{2}-1)^{\frac{5}{2}}+\frac{1}{12}(4x^{2}-1)^{\frac{3}{2}}+C$