Evaluating $\int x \cdot \sqrt{x-1} \cdot \mathrm{d}x$ using substitution

Question

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?

Answer 1

You made just a small error: $\int u^2 du=\frac{u^3}{3}+C$. Therefore following your approach, $$\begin{align}\int x \cdot \sqrt{x-1} dx&=2\int (u^4 + u^2) \cdot \mathrm{d}u=\frac{2u^5}{5}+\frac{2u^3}{3}+C\\&=\frac{2(x-1)^{5/2}}{5}+\frac{2(x-1)^{3/2}}{3}+C\\ &=\frac{2(3x^2-x-2)\sqrt{x-1}}{15}+C.\end{align}$$

Answer 2

$$\int u^2\,\mathrm du=\frac13u^3.$$

Answer 3

Maybe it is easier to take $x-1=y$, or equivalently $x = y+1$, so that $$ \int x\sqrt{x-1} = \int (y+1) \sqrt{y} = \int y^{3/2} + y^{1/2} = \frac{2}{5} y^{5/2} + \frac{2}{3} y^{3/2} = \frac{2}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} $$

Answer 4

Because

$$2 \cdot \int (u^4 + u^2) \mathrm{d}u = 2 \cdot \left(\frac{1}{5} u^5 + \color{red}{\frac{1}{3} u^3}\right) + c = \frac{2}{5} u^5 + \color{red}{\frac{2}{3}u^3} + c$$

$$ = \frac{2 (\sqrt{x-1})^5}{5} + \color{red}{\frac{2\sqrt{(x-1)^3}}{3}} + c$$