integration by substitution?

Question

use x=u^2
integrate between limits 36 and 0
integral 1/(x^1/2(2+x^1/2))dx
and show it equals: ln 16
thanks

Answer 1

Let x = u²
dx = 2u du
I = 2 ∫ u / [ (u) (u + 2) ] du (limits 0 to 6)
u / (u)(u + 2) = A / u + B / (u + 2)
u = A(u + 2) + Bu
1 = A + B
2A = 0
A = 0 , B = 1
I = 2 ∫ 1 / (u + 2) du
I = 2 [ log (u + 2) ]
i = 2 [ log 8 - log 2 ]
I = 2 log 8/2
I = 2 log 4
I = log 4²
I = log 16

Answer 2

integral from 0 to 36 1/(x^1/2(2+x^1/2))dx

let x=u^2
dx=2u du

also need to change the end points from x land to u land. so plug in x=0 and x=36 into x=u^2 to find your new end points.
0=u^2
0=u
36=u^2
6=u

Now substitute everything back into your original integral.
integral from 0 to 36 1/(x^1/2(2+x^1/2))dx
becomes:
integral from 0 to 6 (2u du)/(u(2+u))
the u cancels out from top and bottom
integral from 0 to 6 (2 du)/((2+u))
Bring the 2 out of your integral.
2 (integral from 0 to 6 du/((2+u)) )
rewrite as
2 (integral from 0 to 6 (2+u)^-1 du )
=2 ln(2+u) | (from 0 to 6)

2 ln(2+6)-2 ln(2+0)
2 ln(8)-2ln(2)
2 (ln(8)-ln(2))
2 ln (8/2)
2 ln (4)
ln (4)^2
ln (16)

Answer 3

water's approach makes the most sense, but if you must use the substitution u^2 = x...then...

2udu = dx

int of 2udu / ( u ( u + 2 ) ) which becomes

int of 2du / ( u + 2) -->integrates as 2 ln| u + 2 |

the limts changed as:
x = 0 becomes u = 0
x = 36 becomes u = 6

and we have

2( ln(8) - ln(2) ) = 2(3ln(2) - ln(2)) = 4ln(2) = ln(16)

Answer 4

rewriting the integrand..
integral(from 0 to 36): (x^(-1/2)/(2+x^1/2))dx
let u = (2+x^1/2)
now 2du = 1/2x^(-1/2)dx
thus, 2 integral(0 to 36): du/u
yields 2 ln(2+x^1/2) ] (from 0 to 36)
2 [ln(2+ (sqrt(36)) - ln(2 + 0)]
2[ln(2 + 6) - ln(2)
2[ln(8) - ln(2)]
2 ln(8/2)
2 ln(4)
ln(4)^2
ln(16)

Wheres my 10 points!

Answer 5

let x=u^2
dx=2udu.
original equation becomes
2udu/(u^(2*1/2)*(2+u^(2*1/2)))
=2udu/(u^(2+U))
= (2u)du/(u^2+u^u)
further solve do it yourself.
Keep trying .

Answer 6

a nice question and nice approach ok try in the way u are going