Page 220 - Calculus Demystified

CHAPTER 7
Methods of Integration
Thus we wish to write the integrand as the sum of a factor with denominator 207
2
(x + 2) and another factor with denominator (x + 1). The correct way to do
this is
x x A Bx + C
= = + .
3
2
2
2
x + 2x + x + 2 (x + 2)(x + 1) x + 2 x + 1
We put the right-hand side over a common denominator to obtain
2
x A(x + 1) + (Bx + C)(x + 2)
= .
2
3
2
3
x + 2x + x + 2 x + 2x + x + 2
Identifying numerators leads to
2
x = (A + B)x + (2B + C)x + (A + 2C).
This equation must be identically true, so we find (identifying powers of x) that
A + B = 0
2B + C = 1
A + 2C = 0
Solving this system, we find that A =−2/5, B = 2/5, C = 1/5. So

xdx −2/5 (2/5)x + (1/5)
= dx + dx
2
3
2
x + 2x + x + 2 x + 2 x + 1

−2 1 2x 1 dx
= log |x + 2|+ dx + dx
2
2
5 5 x + 1 5 x + 1
−2 1 2 1
= log |x + 2|+ log |x + 1|+ arctan x + C.
5 5 5
You Try It: Calculate the integral
1 dx
.
2
3
0 x + 6x + 9x
You Try It: Calculate the integral

dx
.
3
x + x
7.3 Substitution
Sometimes it is convenient to transform a given integral into another one by means
of a change of variable. This method is often called “the method of change of
variable” or “u-substitution.”