Problem with simple integral I'm trying to solve this simple integral:
$$\frac12 \int \frac{x^2}{\sqrt{x + 1}} dx$$
Here's what I have done so far:


*

*$\displaystyle t = \sqrt{x + 1} \Leftrightarrow x = t^2 - 1 \Rightarrow dx = 2t dt$

*$\displaystyle \frac12 \int \frac{x^2}{\sqrt{x + 1}} dx = \int \frac{t (t^2 - 1)^2}t dt$

*$\displaystyle \int (t^2 - 1)^2 dt = \frac15 t^5 - \frac23 t^3 + t + C$

*$\displaystyle \frac15 t^5 - \frac23 t^3 + t + C = \frac15 \sqrt{(x + 1)^5} - \frac23 \sqrt{(x + 1)^3} + \sqrt{x + 1} + C$
WolframAlpha tells me steps 1 and 3 are right so the mistake must be somewhere in steps 2 and 4, but I really can't see it.

There is a (slightly) more obvious way of solving it: rewrite the numerator as $x^2+1-1$ and then the whole integral as a sum of two integrals:
$$
\int \frac{(x^2-1)dx}{\sqrt{x+1}} + \int \frac{dx}{\sqrt{x+1}}
$$ 
The second integral is easy, the first one is 
$$
\int \frac{(x^2-1)dx}{\sqrt{x+1}} =\int \frac{(x+1)(x-1)dx}{\sqrt{x+1}}=\int x \sqrt{x+1}dx-\int \sqrt{x+1}dx
$$
The second integral is also easy, and the integrand in the first one should be rewritten and $(x+1-1)\sqrt{x+1}$ and the rest is easy. This is a bit too straightforward, I admit.  
