5. $int_{0}^{sqrt{7}} sqrt3{1 + x^{2}} cdot xdx$

Answer:

$\frac{3}{4}(2^{\frac{4}{3}} - 1)$

Explanation:

Step1: Use substitution

Let $u = 1 + x^{2}$, then $du=2xdx$. When $x = 0$, $u = 1$; when $x=\sqrt{1}$, $u = 2$.

Step2: Rewrite the integral

$\int_{0}^{\sqrt{1}}\sqrt[3]{1 + x^{2}}\cdot xdx=\frac{1}{2}\int_{1}^{2}u^{\frac{1}{3}}du$

Step3: Apply power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$. So, $\frac{1}{2}\int_{1}^{2}u^{\frac{1}{3}}du=\frac{1}{2}\left[\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{1}^{2}$

Step4: Simplify the expression

$\frac{1}{2}\times\frac{3}{4}\left[u^{\frac{4}{3}}\right]_{1}^{2}=\frac{3}{8}(2^{\frac{4}{3}}-1^{\frac{4}{3}})=\frac{3}{4}(2^{\frac{4}{3}} - 1)$