This document discusses integration by substitution. It provides an example of recognizing a composite function and rewriting the integral in terms of the inside and outside functions. Specifically, it shows rewriting the integral of (x2 +1)2x dx as the integral of the outside function (x2 + 1) with the inside function (x) plugged in, plus a constant. It then provides additional practice problems applying the technique of substitution to rewrite integrals in terms of u-substitutions.

1. Integration by Substitution

2. Recognizing the “Outside-Inside” Pattern
∫(x
2
+1) 2x dx
2
From doing derivatives we need to recognize the integrand above
is a composite function from the “derivative of the outside times
the derivative of the inside” (chain rule).
3
1 2
= ( x + 1) + C
3
“+ C” since this is an
indefinite integral

3. Think of this function as 2 functions: f(x) and g(x)
f ( x) = x
g ( x) = x + 1
2
2
As a composite function then:
(
)
f ( g ( x) ) = x + 1
outside
2
2
inside
Now look at the original integral:
∫(x
2
+1) 2x dx
f(g(x))
2
g’(x)

4. Read this as “the
antiderivative of the outside
function with the inside
function plugged in…plus
C”

6. Let’s Practice !!!
3x 2 sin x 3 dx =
∫
3x 2 sin x 3 dx =
∫
du
Let u = x3
du = 3x2
sin u
∫ sin u du =
− cosu + C =
−cos x + C
3

7. More Practice !!!
∫x
1/4
3
x + 2 dx =
4
Let u = x4 + 2
(x
4
x 4 + 2 x 3 dx = 1
4
u
du
∫
1
4
du = 4 x3
4
∫
u du =
3
2
3
2
1u
1 2u
∫ u du = 4 3 + C = 4 3 + C =
2
+ 2)
6
1
2
3
2
+C =
(x
4
+ 2)
6
3
+C

8. Here are some problems for
you to work on!!!
dx
1. ∫
=
2
1+ 3x
1 2 
2. ∫  sin π x ÷dx =
x

3. ∫ sin 2 x cos x dx =
e x
4. ∫
dx =
x
43
5
5. ∫ t 3− 5t dt =

9. Less Apparent
Substitution
1. ∫ x 2 x −1 dx =
∫
x 2 x −1 dx =
(u + 1)
Let u = x – 1
du = dx
x=u+1
x2 = (u + !)2
7
2
∫ ( u +1)
∫(
5
2
u
2
du
u du =
∫ ( u +1)
2
1
2
u du =
3
1
 5
2
 u 2 + 2u 2 + u 2 ÷du =
u + 2u +1 u du = ∫


)
3
2
1
2
7
2
5
2
3
2
2 ( x −1)
4 ( x −1)
2 ( x −1)
2u 2 ×2u 2u
+
+
+C =
+
+
+C
7
5
3
7
5
3

10. Less Apparent
Substitution
1. ∫ x 2 x −1 dx =
∫
x 2 x −1 dx =
(u + 1)
Let u = x – 1
du = dx
x=u+1
x2 = (u + !)2
7
2
∫ ( u +1)
∫(
5
2
u
2
du
u du =
∫ ( u +1)
2
1
2
u du =
3
1
 5
2
 u 2 + 2u 2 + u 2 ÷du =
u + 2u +1 u du = ∫


)
3
2
1
2
7
2
5
2
3
2
2 ( x −1)
4 ( x −1)
2 ( x −1)
2u 2 ×2u 2u
+
+
+C =
+
+
+C
7
5
3
7
5
3