# Integrate this equation by parts

1. Dec 31, 2012

### Xetman

1. The problem statement, all variables and given/known data
Integrate ettdt using integration by parts.

2. The attempt at a solution

Seems like a very easy problem; however, I just started learning integration by parts-- didn't understand this.

So the book's method:

u=t dv=etdt
du=dt v= et

uv-∫vdu=tet-∫etdt=tet-et+C.

My method:

u=et dv=tdt
du=etdt v=t2/2

uv-∫vdu=ett2/2-∫(t2/2)etdt

My method seems wrong, so can anyone please answer me why my method won't work?
Is it just about assigning the correct values to the correct variables?

2. Dec 31, 2012

### lurflurf

Your method is not wrong, but it is not helpful. Integration by parts offers several options. We want to replace our integral by a simpler one. Your integral is more complicated. ∫te^t dt can be traded for ∫t^2 e^t dt or ∫e^t dt, the former is worse the latter better. As you practice you will see what choice improves the situation and what does not.

3. Dec 31, 2012

### MrWarlock616

can you please explain how ? I'm confused..

4. Dec 31, 2012

### tiny-tim

Hi MrWarlock616!

tet is in two parts, t and et

you can push et up, and t down …

then et stays as et, and t becomes 1, then 0​

of you can push et down, and t up …

then et stays as et, and t becomes t2/2, then t3/6, then …​

the idea is to push in the direction that makes one of them disappear

5. Jan 1, 2013

### MrWarlock616

This must a new concept. I'm not familiar with the push up and down..but thanks!

6. Jan 1, 2013

### Curious3141

The thing to keep in mind is that for elementary integration by parts problems, the tricky term is the integral one is left with. From the original integrand (which is the product of two terms), one term gets differentiated, while the other term gets integrated. Do that in your head, and if the product between the two resulting terms can now be integrated again more easily, then it's the right approach.

Writing it out makes it seem more complicated than it is!

7. Jan 1, 2013

### SithsNGiggles

The pushing "up and down" means "integrating and differentiating," respectively. I imagine tiny_tim was thinking of "pushing down" the degree of the "t" part of the original integrand, and the resulting integral (the v du part) would simply be et, as opposed to the one you got using your method.