Integration by Parts

[suicoted]

I tried this question, I got stuck.

S e^(s root of x) dx (S is the integral symbol) (e is to the power of square root x)

I used the rule: S u dv = uv - S v du (S is the integral symbol)

I went let u = e^(sroot x) -> du = 1e(sroot x)/2(sroot x) dx

dv = dx -> v = x

= xe^(sroot of x) - 1/2 S (x/sroot of x) e^(s root of x) dx

Now what?!
I could let use this rule again and let a new variable w = something, to get dw, and dz and z, but what should I make it equal to?

Thanks so much if you guys can help .

 

[soroban]

Hello, suicoted!

. . ∫ e^√x dx

I recommend this substitution first . . .

Let . w = x^1/2 . ---> . x = w² . ---> . dx = 2w dw

Substitute: . 2 ∫ w e^w dw

. . and now try "by parts" . . .

 

[suicoted]

Please help, I'm still lost and confused?

Where did x = w^2 come from?


What do I do after? I'm still new at integration by parts.

 

[stapel]

The tutor was using this substitution to get rid of the radical on the variable. Just add this to your catalog of integration "tricks".

While differentiation is largely formulaic, integration is often troublesome, so watching practiced experts -- and soroban is a great one -- is an excellent way to pick up the "tricks of the trade". Don't worry about being "new"; we all were, once! :wink:

Eliz.

 

[tkhunny]

We have a little advantage on this one, since de^w/dw = e^w

2 ∫ w e^w dw = 2 ∫ w d(e^w) = 2*(w*e^w - ∫ e^w dw)

Is it looking easier, yet?