1. Integration by Parts
If u and v are functions of x and have
continuous derivatives, then

2. Steps for Integration By Parts IBP
1. Express the Integrand as a product of two functions.
2. Find u using the LIPET rule, and name the other functions times dx, dv.
3. Find du by differentiation and v by integration.
4. Substitute u, du, v, & dv into the IBP formula and evaluate the integral of vdu.

3. LIPET is your new order of selection of u for IBP
Logarithm I
Inverse Trig Functions P
Polynomials E
Exponential Functions T
Trig Functions

4. VIDEO TUTORIALS (Allow few second to load)

5. Summary of Common Integrals Using
Integration by Parts
1. For integrals of the form
Let u = xn and let dv = eax dx, sin ax dx, cos ax dx
2. For integrals of the form
Let u = lnx, arcsin ax, or arctan ax and let dv = xn dx
3. For integrals of the form or
Let u = sin bx or cos bx and let dv = eax dx

6. Evaluate u dv u dv u dv u dv To apply integration by parts, we want
to write the integral in the form
There are several ways to do
this.
u dv u dv u dv u dv
Following our guidelines, we choose the first option
because the derivative of u = x is the simplest and
dv = ex dx is the most complicated.

7. u = x
v = ex
du = dx
dv = ex dx
u dv

8. INTERATCTIVE INTEGRATION BY PARTS WORKSHEET (Soln Next Page)

9. INTERACTIVE INTEGRATION BY PARTS WORKSHEET
Allow few minutes to load. Click on problem for detailed solutions

10. Since x2 integrates easier than
ln x, let u = ln x and dv = x2
u = ln x
dv = x2 dx

11. Repeated application of integration by parts
u = x2
v = -cos x
du = 2x dx
dv = sin x dx
Apply integration by
parts again.
u = 2x
du = 2 dx
dv = cos x dx
v = sin x

12. Repeated application of integration by parts
Neither of these choices for u
differentiate down to nothing,
so we can let u = ex or sin x.
Let’s let u = sin x
v = ex
du = cos x dx
dv = ex dx
u = cos x
v = ex
du = -sinx dx
dv = ex dx

14. Guidelines for Integration by Parts
Try letting dv be the most complicated
portion of the integrand that fits a basic
integration formula. Then u will be the
remaining factor(s) of the integrand.
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.