Page 212 - Calculus Demystified

CHAPTER 7
Methods of Integration
If we had instead selected u = cos x and dv = xdx then we would have found 199
2
that v = x /2 and du =− sin xdx and the new integral
2
x
vdu = (− sin x) dx
2
is more complicated.
EXAMPLE 7.2
Calculate the integral

2 x
x · e dx.
SOLUTION
Keeping in mind that we want to choose u and v so as to simplify the integral,
x
2
we take u = x and dv = e dx. Then

u(x) = x 2 du = u (x) dx = 2xdx
x

v(x) = e x dv = v (x) dx = e dx
Then the integration by parts formula tells us that

2 x 2 x x
x e dx = udv = uv − vdu = x · e − e · 2xdx. (∗)

We see that we have transformed the integral into a simpler one (involving
x
2
x
x · e instead of x · e ), but another integration by parts will be required. Now
x
we take u = 2x and dv = e dx. Then

u(x) = 2x du = u (x) dx = 2 dx
x

v(x) = e x dv = v (x) dx = e dx
So equation (∗) equals

2 x 2 x
x · e − udv = x · e − u · v − vdu

2 x x x
= x · e − 2x · e − e · 2 dx
2
x
x
x
= x · e − 2x · e + 2e + C.
We leave it to the reader to check this last answer by differentiation.
You Try It: Calculate the integral

2
x log xdx.