Example ∫
x
x^2 − 3 x+ 2
dx=
∫
x
(x− 1 )(x− 2 )
dx
(
factorise
denominator
)
( 45
➤
)
=
∫ [
2
x− 2
−
1
x− 1
]
dx
(
split into
partial fractions
)
( 62
➤
).
= 2
∫
dx
x− 2
−
∫
dx
x− 1
(isolate integrals) ( 258
➤
).
=2ln(x− 2 )−ln(x− 1 )+C(using substitutions)
You can see the skill required at each step – if any steps still puzzle you go back to the
appropriate paragraph. For a similar example using completing the square see the solution
to the review question
Given a rational function of the type
ax+b
cx^2 +dx+e
we may need to combine a number of methods – such as completing the square and
substitution for example. This may require some manipulation of the integrand. Also we
may not be told which method to use to integrate it, so we have to be able to spot the
best approach. There are three methods we can employ, separately, or in combination:
- Substitution
- Completing the square
- Partial fractions
We can use substitution immediately if the top is proportional to the derivative of the
bottom, using the result
∫
f′(x)dx
f(x)
=lnf(x)+C
If it is not, then check whether the denominator factorises – if it does, then we can use
partial fractions. If the bottom doesn’t factorise, we can always rewrite the numerator to
make it equal to the derivative and an additional constant, that is:
ax+b≡
a( 2 cx+d)
2 c
−
ad
2 c
+b
(Confirm this result to check that your algebra is up to scratch!) Then
ax+b
cx^2 +dx+e
=
a
2 c
(
2 cx+d
cx^2 +dx+e
)
+
(
b−
ad
2 c
)
1
cx^2 +dx+c
The first part can now be done by substitutingu=cx^2 +dx+e, which leaves us with
the problem of integrating something of the form
∫
dx
cx^2 +dx+e
If the bottom doesn’t factorise then we will have to use completing the square.