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

ThePizzaGuy

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Feb 23, 2016
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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 :(
 
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'.

xx1dx\displaystyle \displaystyle{\int \dfrac{x}{\sqrt{x-1}}dx}

u2=x12udu=dx\displaystyle u^2 = x - 1 \to 2u*du = dx

Now you have:

u2+1u(2u)du\displaystyle \displaystyle{\int \dfrac{u^2 + 1}{u}(2*u) du}

Now continue....
 
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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 t=x1\displaystyle t=x-1 then t+1tdt\displaystyle \displaystyle\int {\frac{{t + 1}}{{\sqrt t }}dt}
 
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