Class Notes – The “Box Method” of Integration
The “Box Method” is a guess & check method of finding integrals of quantities raised to powers, or of
products and quotients of those quantities. The “Box Method” replaces the substitution method that is more
common in textbooks, and we use it because it allows you to integrate much more quickly. After the AP test,
we will learn the more common substitution method.
∫ ( 3x − 2 )
Example 1
2
dx
We know from the power rule, ( 3x − 2 ) has to come from ( 3x − 2 ) , a power one higher than the
2
3
exponent “2” on ( 3x − 2 ) . We will therefore guess that the value of our integral is
2
(1)
← This is our ANSWER and all that is
left is to take the derivative of our answer to
solve for the
.
( 3x − 2 )3 + C
We check by taking the derivative with the “Package Rule”.
Thus, the derivative of our ANSWER is
(2)
3
( 3x − 2 )2 i3 = 9 ( 3x − 2 )2
We solve for the
by setting the equation in (2) above equal to the quantity under the
original problem.
So,
9
=1
1
=
9
in equation (1) with the fraction
∫ x ( 5x
Example 2
2
3
)
3
+ 4 dx 
1
, gives us our final ANSWER:
9
This is the integral of a product – we call this
a parent/child function. The “most
complicated” part of the product is the
parent and the other part is the child
Using the Box Method, we integrate the Parent. The parent must come from
(1)
in the
( 3x − 2 )2 = ( 3x − 2 )2
9
Replacing the
∫
( 5x
3
+4
)
4
+C
← This is our ANSWER and all that is
left is to take the derivative of our answer to
solve for the
.
1
( 3x − 2 )3 + C
9
We check by taking the derivative with the “Package Rule”.
Thus, the derivative of our ANSWER is
(2) 4
( 5x
3
)
(
3
+ 4 i15x 2 =
We solve for the
60x 2 5x 3 + 4
)
3
by setting the equation in (2) above equal to the quantity under the
original problem.
(
60x 2 5x 3 + 4
So,
60
Replacing the
3
(
= x 2 5x 3 + 4
)
x 2 + 1dx
in the
3
=1
1
=
60
in equation (1) with the fraction
∫x
Example 3
)
∫
(
1
1
5x 3 + 4
, gives us our final ANSWER:
60
60
)
4
+C
← This is another parent/child function
1
We “guess” that the parent ( x 2 + 1) 2 comes from
(x
(1)
(
3
2
(2)
2
)
3
+ 1 2 + C ← and take the derivative of this to solve for
)
1
x 2 + 1 2 i2x = 3x
We solve for the
(
)
x2 + 1
1
2
by setting the equation in (2) above equal to the quantity under the
original problem.
So, 3x
(x
2
)
+1
3
Replacing the
1
2
= x x2 + 1
=1
1
=
3
in equation (1) with the fraction
1
, gives us our final ANSWER:
3
3
1 2
x +1 2 + C
3
(
)
∫
in the