Find the indefinite integral of x^5 / sqrt(x^2 + 1).

Step-by-step solution

1. 1

   Exploit the odd power of $x$. Because the exponent $5$ is odd, one factor of $x$ can be paired with $dx$ to form $du$, and the remaining $x^{4}$ is an even power that can be written entirely in terms of $x^{2}$. This is the standard reason $u=x^{2}+1$ works here but would not for an even power.

2. 2

   Set up the substitution.

   $u=x^{2}+1\ \Longrightarrow\ du=2x\,dx\ \Longrightarrow\ x\,dx=\tfrac12\,du,\qquad x^{2}=u-1$

   Then

   $x^{5}dx=\left(x^{2}\right)^{2}\cdot x\,dx=(u-1)^{2}\cdot\tfrac12\,du$

3. 3

   Rewrite the whole integral in $u$.

   $\int\frac{x^{5}}{\sqrt{x^{2}+1}}dx=\frac12\int\frac{(u-1)^{2}}{\sqrt u}\,du=\frac12\int\left(u^{3/2}-2u^{1/2}+u^{-1/2}\right)du$

   Expanding $(u-1)^2=u^2-2u+1$ and dividing each term by $u^{1/2}$ lowers every exponent by $\tfrac12$.

4. 4

   Integrate each power with the rule $\int u^{p}du=\frac{u^{p+1}}{p+1}$.

   $\int u^{3/2}du=\tfrac25u^{5/2},\qquad \int u^{1/2}du=\tfrac23u^{3/2},\qquad \int u^{-1/2}du=2u^{1/2}$

5. 5

   Combine and clear the outer one-half.

   $\frac12\left(\frac25u^{5/2}-\frac43u^{3/2}+2u^{1/2}\right)=\frac15u^{5/2}-\frac23u^{3/2}+u^{1/2}$

6. 6

   Substitute back and check.

   $\int\frac{x^{5}}{\sqrt{x^{2}+1}}dx=\frac15\left(x^{2}+1\right)^{5/2}-\frac23\left(x^{2}+1\right)^{3/2}+\left(x^{2}+1\right)^{1/2}+C$

   Differentiating this numerically at $x=1.3$ gives $2.263813$, and the integrand there is $2.263813$ ✓.