Integrate via substitution: $x^2\sqrt{x^2+1}\;dx$ 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.

I have used the trigonometric substitution $x=tan\theta$ and proceeded to get the same answer as @Aman Kushwaha got via completing the square and @user773458 got via hyperbolic trig identies. Below is my steps for reference:
$$
\int (x^2-\sqrt{x^2 +1})dx \\
=\int tan^2\theta \sqrt{tan^2 \theta + 1}sec^2 \theta \;d\theta \\
=\int tan^2 \theta \sqrt{sec^2 \theta}sec^2 \theta \;d\theta \\
=\int tan^2 \theta sec^3 \theta \;d\theta \\
=\int sec^3 \theta(sec^2 \theta - 1) \; d\theta \\
=\int sec^5 \theta - sec^3 \theta \; d\theta \\
=\int sec^5 \theta \; d\theta - \int sec^3 \theta \; d\theta \\
=sec^3 \theta \; tan \theta \; - \; 3 \int sec^3 \theta \; tan^2 \theta \; d\theta - \Big(sec \theta \; tan \theta \; - \; \int sec \theta \; tan^2 \theta \; d\theta\Big) \\
\therefore 4 \int sec^3 \theta \; tan^2 \theta \; d\theta = sec^3 \theta \; tan \theta \; - \Big(sec \theta \; tan \theta \; - \; \int sec \theta \; tan^2 \theta \; d\theta\Big) \\
\therefore \int sec^3 \theta \; tan^2 \theta \; d\theta = \frac{1}{4} \; \bigg(sec^3 \theta \; tan \theta \; - \Big(sec \theta \; tan \theta \; - \; \int sec \theta(sec^2 \theta - 1)\; d\theta\Big)\bigg) \\
= \frac{1}{4} \; \bigg(sec^3 \theta \; tan \theta \; - \Big(sec \theta \; tan \theta \; - \; \int sec^3 \theta \; d\theta \; - \int sec \theta \; d\theta\Big)\bigg) \\
= \frac{1}{4}\big(sec^3 \theta \; tan \theta \; - \; \frac{1}{2}(\sec \theta \; tan \theta - \ln|sec \theta + tan \theta|)\big) + C\\
=\frac{1}{4}\big(\sqrt{tan^2 \theta + 1}(tan^2 \theta + 1)tan \theta \; - \; \frac{1}{2}(\sqrt{tan^2 \theta + 1}(tan \theta) - \ln | \sqrt{tan^2 \theta + 1} + tan \theta|)\big) + C\\
=\frac{1}{4}\big(\sqrt{x^2 + 1}(x^2 + 1)x \; - \; \frac{1}{2}(\sqrt{x^2 + 1}(x) - \ln | \sqrt{x^2 + 1} + x|)\big) + C\\
=\frac{(2x^3+2x)\sqrt{x^2 +1}-x\sqrt{x^2 + 1} - \ln|\sqrt{x^2 + 1} + x|}{8} + C \\
=\frac{(2x^3 + x)\sqrt{x^2 + 1} - \ln|\sqrt{x^2 + 1} + x|}{8} + C \\
$$
