A study Guide on Integration by Parts
Produced by Bryce Borchers and Keldon Reller
Revised by Learning Center
Integration by Parts is a method used when the integrand is a product of functions.
This method is used because the integral of a product of functions is not the product of
the individual integrals of those functions. The method of integration by parts transforms
the integral into a form much easier to integrate. Integration by parts takes the form
ó f (x ) $g' (x ) dx = f (x ) $g(x ) K ó f ' (x )$g(x ) dx
õ
õ
This equation is also known and more commonly referred to as
ó u $ dv = u$v K ó v $ du
õ
õ
When breaking down an integral into parts the first step is to decide which
function will be (u) and which function will be (dv). There are a lot of different ways to
determine which function is which, but we will teach a specific method, the LIATE
method. The LIATE method determines which function should be (u). Steps of the
LIATE method are as follows.
1) First write out LIATE
L – Logarithmic
I – Inverse Trig
A – Algebraic
T – Trig
E – Exponential
2) Whichever function corresponds with the highest term on the list gets assigned to
the all important (u)
3) Lastly, the other function gets defined as (dv).
Once the functions have been defined as either (u) or (dv) a table can and should
be created to find (du) and (v), needed to complete the equation of integration by parts.
A popular table commonly used looks like this.
u = f(x)
du = f’(x)dx
dv = g’(x)dx
v = g(x)
The next step is to set up the equation ó u $ dv = u$v K ó v $ du
õ
õ
and solve it. Now we will run through some example problems to assess your
understanding of integration by parts.
Sample 1) Solve the integral F ( x )   x cos xdx
Solution
Notes
F ( x )   x cos xdx
ux
du  dx
dv  cos xdx
v  sin xdx
F ( x )  x sin x   sin xdx
F ( x )  x sin x  cos x  C
Sample 2) Solve the integral F ( x )   x 2e x dx
Solution
Notes
F ( x )   x e dx
2 x
u  x2
du  2xdx
dv  e x dx
v  e x dx
F ( x )  x 2e x   2xe x dx
u  2x
du  2dx
dv  e x dx
v  e x dx
F ( x )  x 2e x  2xe x   2e x dx
F ( x )  x 2e x  2xe x  2e x  C
3
Sample 3) Solve the integral F ( x )   xe x dx
1
Solution
Notes
3
F ( x )   xe x dx
1
F ( x )  xe x
3
1
3
  e x dx
1
3
F ( x )  3e 3  e 1   e x dx
1
F ( x )  3e 3  e 1  e 3  e 1  2e 3  2e 1