Math 140, Calculus 2
Table Method of Integration
When one is faced with a large power of x in the integrand this is a simple method of
quickly calculating the integral. This method is only used after one understands the
underlying principle of integration by parts.
When an integral of the form
x
n
f ( x)dx
is encountered the calculation can be tedious and time consuming if one diligently runs
through all the steps of integration by parts. An easier method can be obtained by
calculating several pointed examples and then ascertaining a pattern from this. When
integrating by parts we will use the form:
 udv uv   vdu .
Example: find the indefinite integral
x
2
e x dx .
Let’s first integrate by parts.
2 x
 x e dx = uv   vdu ,
u  x 2 , dv  e x dx
du  2 xdx, v  e x
= x 2 e x  2 e x xdx
We integrate by parts again:
= x 2 e x  2 e x xdx ,
u  x, dv  e x dx
du  dx, v  e x
= x 2 e x  2( xe x   e x dx)
= x 2 e x  2 xe x  2e x ,
As always we add an arbitrary constant. From this result we will develop our simpler
technique. First we make a table with u and dv/dx. On the u side of the table we take the
derivative of u as we move down the column. On the dv/dx side of the table we integrate
as we move down the column. Continue to take the derivative of u until it is zero.
Differentiate
u
dv/dx
x2
ex
Integrate
+
2x
ex
_
2
ex
+
0
ex
Now we take the product of alternate rows and columns as shown above and then sum
them using alternate signs as shown above:
Sum= x 2 e x  2 xe x  2e x .
This result is the same as the integral
x
2
e x dx ,
with the addition of an arbitrary constant.
Let’s do another example. This time we’ll integrate
x
3
sin xdx .
First we find our u and dv.
u  x 3 , dv  sin xdx .
Then we set up our table.
u
dv/dx
x3
sinx
+
3x
-cosx
2
_
6x
-sinx
+
6
cosx
_
0
Now we take our sum:
Sum=  x 3 cos x  3x 2 sin x  6 x cos x  6 sin x ,
plus an arbitrary constant. So our integral is:
x
3
sin xdx =  x 3 cos x  3x 2 sin x  6 x cos x  6 sin x  C .
This can be verified easily.
sinx