Split an unwieldy algebraic fraction into simple pieces, and each piece integrates to a logarithm on its own.
9 min read4 worked examples
Prerequisite
This lesson assumes fluency with decomposing an algebraic fraction into partial fractions — for distinct linear factors, repeated linear factors, and (where relevant) an improper fraction requiring algebraic division first. See Partial Fractions in the Algebra & Functions module for the full decomposition method; here, the focus is purely on integrating the result.
Why partial fractions make integration possible
An algebraic fraction like $\dfrac{7x-3}{(x-2)(x+1)}$ has no obvious antiderivative in its combined form. But once split into partial fractions, $\dfrac{7x-3}{(x-2)(x+1)} \equiv \dfrac{A}{x-2}+\dfrac{B}{x+1}$, each term is now simple enough to integrate directly using the reverse-chain-rule result from substitution:
This follows since $\dfrac{d}{dx}\big[\ln|ax+b|\big] = \dfrac{a}{ax+b}$ — so dividing by the constant $a$ exactly undoes that extra factor. Every distinct linear-factor term in a partial fraction decomposition integrates to a logarithm this way.
Repeated linear factors
When a denominator has a repeated factor, such as $(x+1)^2$, the corresponding partial fraction term takes the form $\dfrac{C}{(x+1)^2}$ — a power of a linear expression, not a reverse-log term. This integrates instead using the power rule (after rewriting as $C(x+1)^{-2}$):
Recognising which rule applies — logarithm for a linear denominator to the power 1, power rule for higher powers — is essential once repeated factors are involved.
Divide the first term's coefficient by the x-coefficient (2) in its denominator.
4
$=-\dfrac57\ln|2x+1|+\dfrac57\ln|x-3|+c$.
Simplify.
▪ solved
Common trap
Forgetting to divide by the coefficient of $x$ when the linear factor isn't simply $x+b$ — the extra $\dfrac1a$ factor from $\int\frac{k}{ax+b}dx=\frac{k}{a}\ln|ax+b|+c$ is easy to miss under exam pressure.
The (x−1)⁻² term integrates via the power rule, not as a logarithm.
▪ solved
The key insight
Only linear factors raised to the power $1$ integrate to logarithms. A repeated factor's squared term, $\dfrac{B}{(x-1)^2}=B(x-1)^{-2}$, is a power function and must be integrated with the power rule, giving $-\dfrac{B}{x-1}$ — mixing this up with the logarithm rule is a very common error.
Worked example 4
Definite integral with partial fractions · Challenge
Find the exact value of $\displaystyle\int_1^2 \dfrac{4x^2-2x-11}{(x+1)(x-3)}\,dx$, given the fraction is improper.
MethodDivide first (improper fraction), decompose the remainder, then integrate
1
Degree of numerator (2) ≥ degree of denominator (2), so the fraction is improper: divide to get $\dfrac{4x^2-2x-11}{(x+1)(x-3)} = 4 + \dfrac{Ax+B\text{(remainder)}}{(x+1)(x-3)}$.
An improper algebraic fraction must be split into a polynomial plus a proper fraction first.
2
Since $(x+1)(x-3)=x^2-2x-3$, and $4(x^2-2x-3)=4x^2-8x-12$: remainder $=(4x^2-2x-11)-(4x^2-8x-12)=6x+1$.
Subtract 4 times the denominator from the numerator.
3
So the integrand is $4+\dfrac{6x+1}{(x+1)(x-3)}$. Let $\dfrac{6x+1}{(x+1)(x-3)}\equiv\dfrac{P}{x+1}+\dfrac{Q}{x-3}$: at $x=-1$, $-5=-4P\Rightarrow P=\dfrac54$; at $x=3$, $19=4Q\Rightarrow Q=\dfrac{19}{4}$.
Skipping the initial division would leave an improper fraction that cannot be decomposed by the standard method at all — the "+4" polynomial term is not optional, and integrates to a straightforward $4x$ that must not be forgotten in the final answer.
Common mistake
Applying the logarithm rule to every partial fraction term without checking the power of each linear factor. A term like $\dfrac{B}{(x+2)^2}$ must be integrated with the power rule (giving $-\dfrac{B}{x+2}$), not with a logarithm — only genuinely linear (power 1) denominators produce a $\ln|\cdot|$ term.
Check yourself
Find $\displaystyle\int \dfrac{9}{x(x+3)}\,dx$. (Hint: decompose into partial fractions with denominators x and x + 3 first.)
Method
1
$\dfrac{9}{x(x+3)}\equiv\dfrac{A}{x}+\dfrac{B}{x+3}$. At $x=0$: $9=3A\Rightarrow A=3$. At $x=-3$: $9=-3B\Rightarrow B=-3$.