Integral of x/(sqrt(x-1))

[ThePizzaGuy]

I tried substitution u^2 = (x-1), and i tried u=(x-1). I know there's some stupid error I'm making but for the life of me I cannot figure it out

1)
u^2 = x+1

integral of x/sqrt(x-1) = integral of ((u^2)+1)/(u)

2)
integral of ((u^2)+1)/(u) = integral of u^2/u + integral of 1/u

3)
equals u^2/2 + ln(abs(sqrt(x-1)))+C

4)
compare to WolframAlpha answer of (2/3)(sqrt(x-1))(x+2)+C

I can't see what I'm doing wrong

 

[Deleted member 4993]

I tried substitution u^2 = (x-1), and i tried u=(x-1). I know there's some stupid error I'm making but for the life of me I cannot figure it out

1)
u^2 = x - 1

integral of x/sqrt(x-1) = integral of ((u^2)+1)/(u)

2)
integral of ((u^2)+1)/(u) = integral of u^2/u + integral of 1/u

3)
equals u^2/2 + ln(abs(sqrt(x-1)))+C

4)
compare to WolframAlpha answer of (2/3)(sqrt(x-1))(x+2)+C

I can't see what I'm doing wrong

You forgot to convert 'dx' to 'du'.

\(\displaystyle \displaystyle{\int \dfrac{x}{\sqrt{x-1}}dx}\)

\(\displaystyle u^2 = x - 1 \to 2u*du = dx\)

Now you have:

\(\displaystyle \displaystyle{\int \dfrac{u^2 + 1}{u}(2*u) du}\)

Now continue....

 

[pka]

I tried substitution u^2 = (x-1), and i tried u=(x-1). I know there's some stupid error I'm making but for the life of me I cannot figure it out

What about using \(\displaystyle t=x-1\) then \(\displaystyle \displaystyle\int {\frac{{t + 1}}{{\sqrt t }}dt} \)