Integration by Parts
Integration by Parts Theorem
If u and v are of x and have continuous derivatives, then
§ u dv = uv - § v du.

This formula expresses the original integral in terms of another integral. Depending on the choices for u and dv, it may be more advantageous to evaluate the second integral than the original one. Since the choices of u and dv are critical in the integration by parts method, observe the integration by part notes below. Also, you may wonder what an integrand is. An integrand is the thing that you're integrating, or what ever is in front of dx.

Integration by Parts Notes
1.First try letting dv be the most complicated portion of the integrand that fits a basic integration rule. Then u will be the remaining factor(s) of the integrand.
2.Then try letting u be the portion of the integrand whose derivative is a simpler function than u. Then dv will be the remaining factor(s) of the integrand.


Example:

Evaluate § x2 ln x dx.

Solution: You may notice that x2 is more easily integrated than ln x. Moreover, the derivative of ln x is simpler than the integral of ln x. As a result, let dv = x2 dx.
dv = x2 dx ----->      v = § x2 dx = x3/3
u = ln x ------>      du = 1/x dx
Now use Integration by parts which produces:
§ x2 ln x dx = x3/3 ln x - § (x3/3)(1/x) dx

      = x3/3 ln x - 1/3 § x2 dx
      = x3/3 ln x - x3/9 + C

Differentiate this answer and see if you get x2 ln x.