Issue with intregration question 
$$
\int_0^1 \frac{5x}{\sqrt{x^2+1}}dx
$$

Hi having an issue with the question above.
So far, I have
$ u = x^2 + 1$
$ \frac {du}{dx} = 2x $
$ 2xdu = dx$
$\int_2^1 \frac {5x}{\sqrt(x^2 + 1)} 2xdu $
Stuck after this. Cant seem to factorise?
Any help would be appreciated
$\frac 52 \int_1^2 \frac {du}{\sqrt(u)} $
$\frac 52 \int_1^2 u^\frac{-1}{2} $
$\frac 52 \int_1^2 2u^\frac12 $
$\frac 52 (2^2 + 1) - (1^1 + 1) $
$\frac 52 (5 - 2) = 7\frac12 $
Is this right?
 A: If $u = x^2 + 1$ then $du = 2xdx$ and so you get
$$
\int_0^1 \frac{5x}{\sqrt{x^2+1}}dx
 = \frac{5}{2} \int_0^1 \frac{2xdx}{\sqrt{x^2+1}}
 = \frac{5}{2} \int_1^2 \frac{du}{\sqrt{u}}
$$
Can you take it from here?
UPDATE
Following your updated question, here is the rest of the solution
$$
\begin{split}
\int_0^1 \frac{5x}{\sqrt{x^2+1}}dx
 &= \frac{5}{2} \int_1^2 \frac{du}{\sqrt{u}}
  = \frac{5}{2} \int_1^2 u^{-1/2}du\\
 &= \frac{5}{2} \left. 2u^{1/2}\right|_1^2 \\
 &= \frac{5}{2} \times 2 \times \left(\sqrt{2} - \sqrt{1} \right) \\
 &= 5 \left(\sqrt{2}-1 \right)
\end{split}
$$
A: HINT: write your integral in the form $$\frac{5}{2}\int \frac{2x}{\sqrt{x^2+1}}dx$$ and set $$t=x^2+1$$
A: Your mistake is on this step:
$$\frac{du}{dx}=2x \not \Rightarrow 2x~du=dx$$
However, it should be:
$$\frac{1}{2x}~du=dx$$
Then, the $x$ terms should cancel out.

Also, you've made a mistake in the order of the integration limits. Note that:
$$u=0^2+1=1 \qquad \text{where}~~ x=0$$
And:
$$u=1^2+1=2 \qquad \text{where}~~ x=1$$
Therefore, your integration limits should be from $1$ to $2$ in that order.

Combining this, you should obtain:
$$\int_1^2 \frac{5}{2\sqrt{u}}~du$$
Can you continue?
