Question

Solve each of the integrals in Exercises $39–74$. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$\int x^5\sqrt{x^2-1}dx$

Short answer

The value is$\left(\frac{2x^2}{5}+\frac{1}{7}{\left(x^2-1\right)}^2-\frac{1}{15}\right){\left(x^2-1\right)}^{\frac{3}{2}}+C$

Step-by-step solution

Step 1. Given information

An expression is given as$\int x^5\sqrt{x^2-1}dx$

Step 2. Solving integral

Let

$$\begin{gathered} u=x^2 \\ du=2xdx \end{gathered}$$

Now substitute the value and simplify as,

$$\begin{gathered} \int x^5\sqrt{x^2-1}dx \\ =\int \frac{u^2}{2}\sqrt{u-1}du \\ =\frac{1}{2}\int u^2\sqrt{u-1}du \\ =\frac{1}{2}\int 2v^2{\left(v^2+1\right)}^{2}dv \\ =\int \left(v^6+2v^4+v^2\right)dv \\ =\int v^{6}dv+2\int v^{4}dv+\int v^{2}dv \\ =\frac{v^7}{7}+\frac{2v^5}{5}+\frac{v^3}{3}+C \\ =\frac{(\sqrt{u-1}{)}^7}{7}+\frac{2(\sqrt{u-1}{)}^5}{5}+\frac{(\sqrt{u-1}{)}^3}{3}+C \\ =\frac{1}{7}{\left(x^2-1\right)}^{\frac{7}{2}}+\frac{2}{5}{\left(x^2-1\right)}^{\frac{5}{2}}-\frac{1}{3}{\left(x^2-1\right)}^{\frac{3}{2}}+C \\ =\left(\frac{2x^2}{5}+\frac{1}{7}{\left(x^2-1\right)}^2-\frac{1}{15}\right){\left(x^2-1\right)}^{\frac{3}{2}}+C \end{gathered}$$