Indefinite integral with a square root? [calc homework]?

Question

Hey all,

So, I was able to get all my problems but this one. I got:

int(   x / (sqrt(x+5)) dx)

Looks like I gotta use U-substitution.

SO,

Let x+5 = u.
Therefore, deriving this equation,

1 dx = du

So, I replace the dx with du.

I now have

int(   x / (sqrt(u)) du)

--------now what?

I thought maybe I could substitute u for just x, but that wouldn't do anything.

What should I do to solve this problem????


THANKS! =)

The Integral ∫ · Accepted Answer

int( x / (sqrt(u)) du)   and knowing u = x + 5 ====> u - 5 = u 

∫ (u - 5) / (√(u) ]

∫ [ u/√(u) ]  -  [ 5/ √(u) ]

∫ √(u) -   5u^(-1/2)  

u^(3/2) / (3/2)   -  5   * u^(1/2) / (1/2)  +  C 

(2/3)  * u^(3/2)  -  5 * 2  * u^(1/2)  +  C 

(2/3)  * ( x + 5)^(3/2)  -  10 ( x + 5)^(1/2)  +  C

ted s · Answer

with your sub you also have x = u - 5....but I prefer w² = x + 5 ---> [ 2 ][ w² - 5 ] dw

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Indefinite integral with a square root? [calc homework]?

2 Answers