Single Variable Calculus, Chapter 5, Review Exercises, Section Review Exercises, Problem 30

Find the indefinite integral $\displaystyle \int \frac{x^3}{\sqrt{x^2 + 1}} dx$. Illustrate by graphing both the function and its antiderivative (take $C =0$).
If we let $u = x^2 +1$, then $du = 2x dx$, so $\displaystyle x dx = \frac{du}{2}$. Also $x^2 = u - 1$. Thus,

$
\begin{equation}
\begin{aligned}
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \int \frac{x^2}{\sqrt{x^2 + 1}} x dx\\
\\
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \int \frac{u- 1}{\sqrt{u}} \frac{du}{2}\\
\\
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \frac{1}{2} \int \frac{u - 1}{\sqrt{u}} du\\
\\
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \frac{1}{2} \int \frac{u}{u^{\frac{1}{2}}} - \frac{1}{u^{\frac{1}{2}}} du \\
\\
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \frac{1}{2} \left( \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} - \frac{u^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} \right) + C\\
\\
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \frac{1}{2} \left( \frac{u^{\frac{3}{2}}}{\frac{3}{2}} - \frac{u^{\frac{1}{2}}}{\frac{1}{2}} \right) + C\\
\\
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \frac{1}{2} \left( \frac{2u^{\frac{3}{2}}}{3} - 2u^{\frac{1}{2}} \right) + C\\
\\
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \frac{u^{\frac{3}{2}}}{3} - u^{\frac{1}{2}} + C\\
\\
\text{but } c = 0 \text{ ,so we have}\\
\\
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \frac{u^{\frac{3}{2}}}{3} - u^{\frac{1}{2}} + C\\
\\
\int \frac{x^3}{\sqrt{x^2 + 1}} dx &= \frac{\left( x^2 + 1 \right)^{\frac{3}{2}}}{3} - \sqrt{x^2 + 1}
\end{aligned}
\end{equation}
$


Graph the function $\displaystyle f(x) = \frac{\left( x^2 + 1 \right)^{\frac{3}{2}}}{3} - \sqrt{x^2 + 1}$



Graph of antiderivative $\displaystyle f'(x) = \frac{x^3}{\sqrt{x^2 + 1}}$