<s>[INST] Evaluating $\int x \cdot \sqrt{x-1} \cdot \mathrm{d}x$ using substitution I know that my result is wrong. What did I do wrong?
Let $u = \sqrt{x-1}$. Then we find first:
$$ u^2 = x - 1 \Rightarrow x = u^2 + 1$$
$$ \Rightarrow \mathrm{d}x = 2u \cdot \mathrm{d}u $$
So:
$$\int x \cdot \sqrt{x-1} \cdot \mathrm{d}x = \int (u^2 + 1) \cdot u \cdot 2u \cdot \mathrm{d}u  = 2 \cdot \int (u^4 + u^2) \mathrm{d}u$$
$$ = 2 \cdot (\frac{1}{5} u^5 + \frac{1}{2} u^2) + c = \frac{2}{5} u^5 + u^2 + c$$
$$ = \frac{2}{5} (\sqrt{x-1})^5 + x + c$$
Differentiating this does not give me $ x \cdot \sqrt{x-1}$, so it must be wrong. Which step went wrong and why?
 [/INST] Because 
$$2 \cdot \int (u^4 + u^2) \mathrm{d}u = 2 \cdot \left(\frac{1}{5} u^5 + \color{red}{\frac{1}{3} u^3}\right) + c = \frac{2}{5} u^5 + \color{red}{\frac{2}{3}u^3} + c$$
$$ = \frac{2 (\sqrt{x-1})^5}{5} + \color{red}{\frac{2\sqrt{(x-1)^3}}{3}} + c$$
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