Calculus integral?

Question

evaluate the integral between 10 and 5 [x/root(x-1)] dx

Answer 1

I am assuming the 10 is on top and 5 is on bottom, if not just add a minus sign to each line.

Let's let y = x-1

Then:
Int(5 to 10)[x/root(x-1)]dx =

Int(4 to 9)[(y+1)/root(y)]dy =

Int(4 to 9)[root(y) +1/root(y)]dy =

[(2/3)y^(3/2) + 2y^(1/2)](4 to 9) =

(2/3)*27 + 2*3 - ((2/3)8 + 2*2) =

18 + 6 - 16/3 - 4 =

44/3

Answer 2

Integral ( [x/sqrt(x - 1)] dx )

To solve this, we use substitution.

Let u = sqrt(x - 1). Square both sides, to get

u^2 = x - 1.

u^2 + 1 = x, and now, differentiating,
2u du = dx

Note that our integration bounds will also change; currently at 5 to 10, if x = 5, u = sqrt(5 - 1) = 2; if x = 10, u = sqrt(10 - 1) = 3.

Our new bounds are going to be from 2 to 3.

Substituting, we get

Integral ( [u^2 + 1]/u du )

Splitting this up into two fractions,

Integral ( [u^2/u + 1/u] du)

Integral ( [u + 1/u] du )

Integrating,

( (1/2)u^2 + ln|u| ) {evaluated from 2 to 3}

= ( (1/2)3^2 + ln|3| ) - ( (1/2)2^2 + ln|2| )
= ( (1/2)(9) + ln(3) ) - ( (1/2)4 + ln(2) )
= (9/2) + ln(3) - 2 - ln(2)
= (9/2) + ln(3) - 4/2 - ln(2)
= 5/2 + ln(3) - ln(2)

Now, we can use the log property to combine the logs.

= 5/2 + ln(3/2)

Answer 3

Take the definite integral from 10 to 5 of ∫{x/√(x - 1)}dx.

∫{x/√(x - 1)}dx
Let
u = x - 1
du = dx
x = u + 1

∫{x/√(x - 1)}dx = ∫{(u + 1)/√u}du
= ∫{(u^(1/2) + u^(-1/2)}du = (2/3)u^(3/2) + 2u^(1/2)
= (√u) [(2/3)u + 2]

x runs from 10 to 5 so u runs from 9 to 4

(√u) [(2/3)u + 2] | {eval from 9 to 4}
This is backwards from what I would have expected.

= {(√4) [(2/3)4 + 2]} - {(√9) [(2/3)9 + 2]}
= {2*(14/3) - 3*8} = 28/3 - 24 = (28 - 72)/3 = -44/3