Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 30

Find the indefinite integral $\displaystyle \int x^3 \sqrt{x^2 + 1} dx$

If we let $\displaystyle u = x^2 + 1$, then $\displaystyle du = 2x dx$, so $\displaystyle xdx = \frac{du}{2}$. Also, $x^2 = u - 1$. Therefore,


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\begin{equation}
\begin{aligned}

\int x^3 \sqrt{x^2 + 1} dx =& \int x^2 \sqrt{x^2 + 1} xdx
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\int x^3 \sqrt{x^2 + 1} dx =& \int (u - 1) \sqrt{u} \frac{du}{2}
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\int x^3 \sqrt{x^2 + 1} dx =& \frac{1}{2} \int (u - 1) u^{\frac{1}{2}} du
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\int x^3 \sqrt{x^2 + 1} dx =& \frac{1}{2} \int (u^{\frac{3}{2}}- u^{\frac{1}{2}} ) du
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\int x^3 \sqrt{x^2 + 1} dx =& \frac{1}{2} \int u^{\frac{3}{2}} du - \frac{1}{2} \int u^{\frac{1}{2}} du
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\int x^3 \sqrt{x^2 + 1} dx =& \frac{1}{2} \cdot \frac{u^{\frac{3}{2} + 1} }{\displaystyle \frac{3}{2} + 1} - \frac{1}{2} \cdot \frac{u^{\frac{1}{2} + 1} }{\displaystyle \frac{1}{2} + 1} + C
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\int x^3 \sqrt{x^2 + 1} dx =& \frac{1}{2} \cdot \frac{u^{\frac{5}{2}}}{\displaystyle \frac{5}{2}} - \frac{1}{2} \cdot \frac{u^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} + C
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\int x^3 \sqrt{x^2 + 1} dx =& \frac{1}{\cancel{2}} \cdot \frac{\cancel{2} u^{\frac{5}{2}}}{5} - \frac{1}{\cancel{2}} \cdot \frac{\cancel{2} u^{\frac{3}{2}}}{3} + C
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\int x^3 \sqrt{x^2 + 1} dx =& \frac{u^{\frac{5}{2}}}{5} - \frac{u^{\frac{3}{2}}}{3} + C
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\int x^3 \sqrt{x^2 + 1} dx =& \frac{(x^2 + 1)^{\frac{5}{2}}}{5} - \frac{(x^2 + 1)^{\frac{3}{2}}}{3} + C

\end{aligned}
\end{equation}
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