Integration By Parts
Key: Keep track of your signs (+ or -) and remember your integration rules. And becareful with your arithmetic and algebra as well; one mistake could make entire problem real ugly..

The formula for parts is

òu * dv = u * v - òv * du

where u and v are functions. Simply put, this means you'll be integrating a function that can be broken into two simpler functions (u and dv). ex * x is a good example. Your two functions would be ex and x.

x * ex - ò ex * 1 dx.

Since I picked u to be x and dv to be ex, it follows that du is 1 dx and v is ex (the integral of ex). Now integrating what's left inside the integral is simple. The integral of ex dx is ex. So your answer is x * ex - ex.

It's not always so easy. Sometimes you'll have to deploy parts more than once. An example is ex * sin(x). Let's set it up.

òex * sin(x) dx

Let u = sin(x) and dv = ex. Then du = cos(x) dx and v = ex.

òex * sin(x) = sin(x) * ex - ò ex * cos(x)

We still can't integrate that. Do it again. Let u = cos(x) and dv = ex. Then du = -sin(x) dx and v = ex.

òex * sin(x) = sin(x) * ex - (cos(x) * ex - ò ex * -sin(x) )

Don't be afraid to use plenty of parentheses to help you keep your signs straight. Cleaning up with some Algebra also helps

òex * sin(x) = ex (sin(x) - cos(x)) - ò ex * sin(x)

Now we have something like A = B - A, or 2A = B, or A = B / 2. It's just more Algebra; don't make anything more of it.

òex * sin(x) = ex (sin(x) - cos(x))

2
+ C
It is important to carry the original setup all the way through the problem. That way you will know when you have A's on both sides of the equation. An alternate way of doing this is tabular integration. I recommend the above way (non-tabular) based on my experiences if for no other reason.

u and derivatives * v and integrals
    ex
x ¾ *+ ® ex
¾ *- ® ex
¾ *+ ® ex
¾ *- ® ex
 
Probable Question: "How do I know what should be u and what should be v?"
Find things that are easy to differentiate (take the derivative of) and possibly go to 0 after a few times. Make that u. Make the other thing v. Just think of it in an easy way.

That's integration by parts. Once again, only two different types of problems to recognize. Once you've recognized if it repeats or just goes through once you've already won half the battle. Things that repeat and are cyclical will be in the trigonometric (including the hyperbolics) and ex families, things that repeat but go to 0 will be your xn types.