question evaluate the indefinite integral given below. ∫(x^5 / √(x^2 - 2)) dx provide your answer below.

Explanation:

Step1: Use substitution

Let $u = x^{2}-2$, then $x^{2}=u + 2$ and $du=2xdx$, $x dx=\frac{1}{2}du$. Also, $x^{5}=x^{4}\cdot x=(x^{2})^{2}\cdot x$. Substituting $x^{2}=u + 2$ and $x dx=\frac{1}{2}du$ into the integral, we get $\int\frac{x^{5}}{\sqrt{x^{2}-2}}dx=\frac{1}{2}\int\frac{(u + 2)^{2}}{\sqrt{u}}du$.

Step2: Expand the numerator

Expand $(u + 2)^{2}=u^{2}+4u + 4$. So the integral becomes $\frac{1}{2}\int\frac{u^{2}+4u + 4}{\sqrt{u}}du=\frac{1}{2}\int(u^{\frac{3}{2}}+4u^{\frac{1}{2}}+4u^{-\frac{1}{2}})du$.

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: $\frac{1}{2}(\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}+4\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+4\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1})+C=\frac{1}{2}(\frac{u^{\frac{5}{2}}}{\frac{5}{2}}+4\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+4\frac{u^{\frac{1}{2}}}{\frac{1}{2}})+C$.

Step4: Simplify the expression

$\frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}+\frac{8}{3}u^{\frac{3}{2}}+8u^{\frac{1}{2}})+C=\frac{1}{5}u^{\frac{5}{2}}+\frac{4}{3}u^{\frac{3}{2}}+4u^{\frac{1}{2}}+C$.

Step5: Substitute back $u = x^{2}-2$

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

Answer:

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