QCE Specialist Mathematics - Unit 4 - Integration techniques
Partial Fractions and Integration by Parts | QCE Specialist Mathematics
Learn QCE Specialist partial fractions and integration by parts, including when each technique is useful.
Updated 2026-05-18 - 4 min read
QCAA official coverage - Specialist Mathematics 2025 v1.4
Exact syllabus points covered
- Use partial fractions for integration involving two distinct linear factors in the denominator, e.g. $\frac{2x-1}{(x+1)(x-2)}$.
- Integrate by parts: $\int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx$.
When substitution does not obviously work, Specialist gives you two more major tools: partial fractions and integration by parts. They solve different problems, so the first skill is recognising which shape you are looking at.
Original Sylligence diagram for specialist integration techniques.
Partial fractions
Partial fractions break one rational expression into simpler rational expressions. For QCE Specialist, the required integration cases involve distinct linear factors in the denominator.
For example:
$ \frac{7}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}. $
Multiplying through by $(x-1)(x+2)$ gives:
$ 7=A(x+2)+B(x-1). $
Now choose useful $x$ values:
If $x=1$,
$ 7=3A,\quad A=\frac73. $
If $x=-2$,
$ 7=-3B,\quad B=-\frac73. $
So:
$ \int \frac{7}{(x-1)(x+2)}\,dx = \frac73\ln|x-1|-\frac73\ln|x+2|+c. $
Before decomposing, check that the numerator degree is less than the denominator degree. If it is not, use polynomial division first. In the QCE distinct-linear-factor case, the setup is usually:
$ \frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}. $
You can find $A$ and $B$ by substitution after clearing denominators, or by comparing coefficients. Substitution is fast when the denominator factors are simple; coefficient comparison is useful when no convenient value removes all but one unknown.
More partial-fraction setups
Three distinct linear factors work the same way:
$ \frac{5x+1}{(x-1)(x+2)(x-3)} =\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}. $
After multiplying through by the denominator, choose $x=1$, $x=-2$ and $x=3$ to isolate $A$, $B$ and $C$. Each term then integrates to a logarithm.
Non-monic factors need one extra bit of care. If a term is:
$ \frac{A}{2x+1}, $
then:
$ \int\frac{A}{2x+1}\,dx=\frac{A}{2}\ln|2x+1|+c. $
The factor $\frac12$ appears because the derivative of $2x+1$ is $2$.
Some course materials go beyond the external-assessment distinct-linear-factor focus and include irreducible quadratic factors. The setup then uses a linear numerator:
$ \frac{px+q}{(x-1)(x^2+4)} =\frac{A}{x-1}+\frac{Bx+C}{x^2+4}. $
The $Bx+C$ is not optional. A constant numerator over the quadratic is too restrictive, because the numerator after clearing denominators may need an $x$ term. After solving for $A$, $B$ and $C$, the quadratic part may split into a logarithmic piece from $\frac{2x}{x^2+4}$ and an arctangent piece from $\frac{1}{x^2+4}$.
Integration by parts
Integration by parts reverses the product rule:
$ \int u\,dv=uv-\int v\,du. $
It is often useful when the integrand is a product and one factor becomes simpler when differentiated.
Common choices for $u$ include logarithmic functions, inverse trigonometric functions and algebraic powers. A common memory guide is LIATE: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential.
The formula is sometimes written in derivative notation:
$ \int u\frac{dv}{dx}\,dx=uv-\int v\frac{du}{dx}\,dx. $
Both versions say the same thing. The practical question is: which part do you want to differentiate, and which part can you integrate cleanly?
When it takes more than one round
Sometimes integration by parts must be used repeatedly. For example, $\int x^2e^x\,dx$ lowers the power from $x^2$ to $2x$, then from $2x$ to a constant. That is a feature, not a failure.
For products like $e^x\sin x$ or $e^x\cos x$, integration by parts may cycle back to the original integral. When that happens, move the repeated integral to the other side and solve algebraically. Do not assume you made a mistake just because the original integral reappears.
Definite integration by parts
Integration by parts also works cleanly for definite integrals:
$ \int_a^b u\,dv=\left[uv\right]_a^b-\int_a^b v\,du. $
For example, to evaluate $\int_0^1 \sin^{-1}x\,dx$, treat the integrand as a product with $1$:
$ u=\sin^{-1}x,\quad dv=dx. $
Then:
$ du=\frac1{\sqrt{1-x^2}}\,dx,\quad v=x. $
So:
$ \int_0^1 \sin^{-1}x\,dx =\left[x\sin^{-1}x\right]_0^1-\int_0^1\frac{x}{\sqrt{1-x^2}}\,dx. $
The remaining integral is a substitution integral with $w=1-x^2$. It equals $1$ on $[0,1]$, so:
$ \int_0^1 \sin^{-1}x\,dx=\frac{\pi}{2}-1. $
Repeated integration by parts is useful for powers times exponentials. For instance, $\int x^2e^x\,dx$ becomes:
$ x^2e^x-2\int xe^x\,dx, $
then integration by parts is used again on $\int xe^x\,dx$. The power drops each time until the integral is direct.
Choosing the technique
Partial fractions are for rational functions after the denominator has been factorised. Integration by parts is for products, especially when one factor simplifies after differentiation. If the integrand is a composite function with its derivative nearby, substitution should still be tried first.