I need to integrate the following using substitution:
$$
\int x^2\sqrt{x^2+1}\;dx
$$
My textbook has a similar example:
$$
\int \sqrt{x^2+1}\;x^5\;dx
$$
They integrate by splitting $x^5$ into $x^4\cdot x$ and then substituting with $u=x^2+1$:
$$
\int \sqrt{x^2+1}\;x^4\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;(u-1)^2\;du\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u^2-2u+1)\;du\\
=\frac{1}{2}\int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}\;du\\
=\frac{1}{7}u^{\frac{7}{2}}-\frac{2}{5}u^{\frac{5}{2}}+\frac{1}{3}u^{\frac{3}{2}}+C
$$
So far so good. But when I try this method on the given integral, I get the following:
$$
\int x^2\sqrt{x^2+1}\;dx\\
=\frac{1}{2}\int \sqrt{x^2+1}\;x\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;\sqrt{u-1}\;du\;(u=x^2+1)\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u-1)^\frac{1}{2}\;du
$$
Here is where it falls down. I can't expand the $(u-1)^\frac{1}{2}$ factor like the $(u-1)^2$ factor above was, because it results in an infinite series. I couldn't prove, but I think any even exponent for the $x$ factor outside the square root will cause an infinite series to result. Odd exponents for $x$ will work, since it will cause the $(u-1)$ term to have a positive integer exponent.
How should I proceed? I don't necessarily want an answer. I just want to know if I'm missing something obvious or if it is indeed above first year calculus level and probably a typo on the question.
Best Answer
One way to proceed is: \begin{align} \int x^2 \sqrt{x^2+1} \, dx &= \frac{1}{2} \int \sqrt{u} \sqrt{u-1} \, du \qquad (u = x^2 + 1) \\ &= - \frac{1}{16} \int \frac{y^8 - 2 y^4 + 1}{y^5} \, dy \qquad (y = \sqrt{u} - \sqrt{u-1}) \\ &= - \frac{y^4}{64} + \frac{1}{64 y^4} + \frac{\log(y)}{8} + C. \end{align}
Here, to change the integration variable from $u$ to $y$, one needs \begin{equation} \frac{dy}{du} = \frac{1}{2} \left( \frac{1}{\sqrt{u}} - \frac{1}{\sqrt{u-1}} \right) = - \frac{y}{2 \sqrt{u} \sqrt{u-1}}, \end{equation} and also the expression of $u$ in terms of $y$, which can be obtained by squaring the both sides of $\sqrt{u-1} = \sqrt{u}-y$ and then solving it for $\sqrt{u}$: \begin{equation} u = \frac{(y^2+1)^2}{4y^2}. \end{equation} So, after the variable transformation from $u$ to $y$, one gets a factor $u(u-1)$, which is rewritten as \begin{equation} u(u-1) = \frac{(y^2+1)^2(y^2-1)^2}{16 y^4}. \end{equation}
Now, after performing the integration, one needs the following substitutions \begin{equation} y = \sqrt{x^2 + 1} - x , \qquad y^4 - \frac{1}{y^4} = - 8 x (2 x^2 + 1) \sqrt{x^2 + 1}, \end{equation} to get the final result in $x$: \begin{equation} \int x^2 \sqrt{x^2+1} \, dx = \frac{1}{8} \left[ x (2x^2+1) \sqrt{x^2+1} + \log(\sqrt{x^2+1} - 1) \right] + C. \end{equation} Note that one can rewrite the log term as: \begin{equation} \log(\sqrt{x^2+1}-x) = - \log(\sqrt{x^2+1}+x) = - \sinh^{-1}(x). \end{equation}