1. Integration by Substitution
If y=g(u) is continuous on an open interval and u=u(x) is a differ-
entiable function whose values are in the interval, then for aand bin the
domain of u
Zb
a
g(u(x))u0(x)dx =Zu(b)
u(a)
g(u)du.
Examples: Evaluate the integrals
1. R1
0
ex
1+exdx. Substitute u= 1 + ex. Then du =exdx. If x= 0, then
u= 1 + e0= 2, and if x= 1, u = 1 + e1= 1 + e. Hence
Z1
0
ex
1 + exdx =Z1+e
2
1
udu = ln u|1+e
2= ln(1 + e)−ln 2 = ln 1 + e
2.
2. R1
0
x
1+x2dx. Substitute u= 1 + x2. Then du = 2xdx. If x= 0, u =
1+02= 1, and if x= 1, u = 1 + 12= 2. Hence
Z1
0
x
1 + x2dx =Z2
1
1
u
du
2=1
2Z2
1
1
udu =1
2ln u|2
1=1
2(ln 2 −ln 1) = 1
2ln 2
3. R3
1
1
x√1+xdx.
Substitute u=√1 + x. Solve for xin order ti find dx. We have x=u2−1
and dx = 2udu. If x= 1, u =√1 + 1 = √2 and if x= 3, u =√1 + 3 = 2.
Hence
Z3
1
1
x√1 + xdx =Z2
√2
1
(u2−1)u2udu = 2 Z2
√2
1
(u−1)(u+ 1)du =
=Z2
√2
1
u−1du −Z2
√2
1
u+ 1du = ln(u−1)|2
√2−ln(u+ 1)|2
√2
1
