Integral involving hyperbolic functions

[DryRun]

Homework Statement
Find \int \frac{x}{\sqrt{2x^2-2x+1}}\,dx

The attempt at a solution

First, i complete the square for the quadratic expression:
2x^2-2x+1=2((x-\frac{1}{2})^2+\frac{1}{4})
\int \frac{x}{\sqrt{2x^2-2x+1}}\,dx=\int \frac{x}{\sqrt 2 \sqrt{(x-\frac{1}{2})^2+\frac{1}{4}}}\,dx

Using substitution: Let (x-\frac{1}{2})=\frac{1}{2}\sinh u

I eventually get this expression, after integrating:
\frac{1}{2\sqrt 2}(\cosh (\sinh^{-1}(2x-1)+\sinh^{-1}(2x-1))

But i don't know how to simplify this any further, without getting a complicated answer. The final answer is: \frac{\sinh^{-1}1}{\sqrt 2}

 

[tiny-tim]

hi sharks!

(you haven't said what the limits are)

I eventually get this expression, after integrating:
\frac{1}{2\sqrt 2}(\cosh (\sinh^{-1}(2x-1)+\sinh^{-1}(2x-1))

But i don't know how to simplify this any further

cosh(sinh^-1x) = √(x² + 1)

(btw, you could have simplified the original integrand by rewriting the numerator as x - 1/2 + 1/2 )

 

[DryRun]

Hi tiny-tim!

Congratulations on your PF award!

Doh! I forgot to include the limits.

\int^1_0 \frac{x}{\sqrt{2x^2-2x+1}}\,dx

If i evaluate over the limits:
\frac{1}{2\sqrt 2}(\cosh (\sinh^{-1}(2x-1)+\sinh^{-1}(2x-1))
Indeed, i get the answer!

I am trying to understand your suggestion:
\cosh (\sinh^{-1}x) = \sqrt{(x^2 + 1)}
It's too complicated to prove? I can't get the R.H.S. from the L.H.S.

Here is where I've reached...
\frac{x^2+x\sqrt{x^2+1}+1}{x+\sqrt{x^2+1}}

 

[tiny-tim]

I am trying to understand your suggestion:
\cosh (\sinh^{-1}x) = \sqrt{(x^2 + 1)}

if y = sinh^-1x

then x = sinhy,

so √(x² + 1) = √(sinh²y + 1) = coshy = cosh(sinh^-1x)

(if you think about it, this is obvious! )

 

[DryRun]

I have also tried to solve this problem using your other suggestion - rewriting the numerator as x - 1/2 + 1/2, and i was able to get the answer.

Thank you for your insightful suggestions, tiny-tim.