Mathematics 9709 · Pure Mathematics 3

When the integral needs partial fractions

Sometimes, after separating, one side is a single fraction whose bottom factorises: one over a product, or one over a quadratic.
You cannot integrate it as it stands. You must first split it into partial fractions.
These are the 9 to 11 mark DE questions. They are worth the most, and the easiest to slip up on.

How to spot it: after separating, a side looks like $\dfrac{1}{(\text{product of linear factors})}$ or $\dfrac{1}{\text{quadratic}}$ $\;\Rightarrow\;$ split it first, integrate to a sum of logs, combine, exponentiate, untangle.

Why you must split it first

You can integrate $\frac{1}{x}$ and $\frac{1}{ax+b}$ straight to logs, but $\frac{1}{x(300-x)}$ is not in that shape.
Partial fractions (the Ch7 method, cover-up or equating coefficients) rewrites it as a sum of those easy pieces, and each one becomes a log.

The chain: decompose → combine logs → exponentiate → untangle

Worked example

Textbook 10.3: $\dfrac{dx}{dt}=\dfrac{x(300-x)}{600}$ , $x=50$ when $t=0$

A population-style model: the variables are almost split already, so all the work is in the $x$ -fraction.

Separate (keep the 600 on the right, simpler): $\dfrac{1}{x(300-x)}\,dx=\dfrac{1}{600}\,dt$ .
Decompose (Ch7): $\dfrac{1}{x(300-x)}=\dfrac{1}{300}\Big(\dfrac{1}{x}+\dfrac{1}{300-x}\Big)$ .
Integrate each side. Note $\int\frac{1}{300-x}\,dx=-\ln(300-x)$ (chain rule sign): $\tfrac{1}{300}\big[\ln x-\ln(300-x)\big]=\tfrac{1}{600}t+C$ . Modulus dropped since $0<x<300$ .
Combine logs with $\ln p-\ln q=\ln\frac{p}{q}$ , and tidy (multiply by 300): $\ln\dfrac{x}{300-x}=\tfrac12 t+C$ .
Find $C$ : at $x=50,\;t=0$ , $C=\ln\frac{50}{250}=\ln 0.2$ .
Exponentiate, then untangle $x$ (cross-multiply, collect $x$ ): $\dfrac{x}{300-x}=0.2\,e^{\frac12 t}$ $\;\Rightarrow\;$ $x\big(1+0.2\,e^{\frac12 t}\big)=60\,e^{\frac12 t}$ .

Answer

x=\dfrac{60\,e^{\frac12 t}}{1+0.2\,e^{\frac12 t}}

Where most students freeze is that last untangle: getting the lone $x$ out of a fraction. Cross-multiply, gather every $x$ term, factor $x$ out, divide.

Common mistake

Three classic slips on this type:

Integrating the fraction before splitting it (there is no clean integral until you split it).
Sign error when combining logs, especially the $-\ln(300-x)$ from the $(300-x)$ piece.
Leaving the answer in $\ln$ form when “ $x$ in terms of $t$ ” was asked. You must exponentiate and get $x$ on its own.

When the question names the technique

Worked example

9709/31 May–Jun 2021 Q10: $\dfrac{dx}{dt}=x^2(1+2x)$ , $x=1$ when $t=0$ ; $t$ in terms of $x$ [11]

The question says “using partial fractions”, so separate with the $x$ -fraction on the left: $\dfrac{1}{x^2(1+2x)}\,dx=dt$ .

Decompose (repeated factor $x^2$ needs both $\frac{A}{x}$ and $\frac{B}{x^2}$ ): $\dfrac{1}{x^2(1+2x)}=\dfrac{1}{x^2}-\dfrac{2}{x}+\dfrac{4}{2x+1}$ .
Integrate: $t=-\dfrac{1}{x}-2\ln x+2\ln(2x+1)+C$ ( $\int\frac{4}{2x+1}=2\ln(2x+1)$ ).
Find $C$ : at $x=1,\;t=0$ , $0=-1-0+2\ln 3+C$ , so $C=1-2\ln 3$ .
Combine logs into one: $2\ln(2x+1)-2\ln x-2\ln 3=2\ln\dfrac{2x+1}{3x}$ .

Answer

t=1-\dfrac{1}{x}+2\ln\!\dfrac{2x+1}{3x}

Checked. Because it asked for $t$ in terms of $x$ , you stop at this $\ln$ form; no exponentiating needed.

Your turn— tap to reveal the worked answer (9709/32 Feb–Mar 2022 Q9 (shorter warm-up))

The $(x+1)(3x+1)$ bottom from the previous topic needs partial fractions too, with $A=-\tfrac12,\;B=\tfrac32$ . Good two-factor practice before the $[11]$ -mark one above. It is worked in full under “Solve a given DE”. The exact value comes out $\sqrt5/2$ .

Answer

y=\dfrac{\sqrt5}{2}\ \text{ at }\ x=3

PreviousSolve a given DE and rearrange to the requested form NextForming a DE from a rate statement, and interpreting the solution

When the integral needs partial fractions

Sometimes, after separating, one side is a single fraction whose bottom factorises: one over a product, or one over a quadratic.
You cannot integrate it as it stands. You must first split it into partial fractions.
These are the 9 to 11 mark DE questions. They are worth the most, and the easiest to slip up on.

Why you must split it first

You can integrate $\frac{1}{x}$ and $\frac{1}{ax+b}$ straight to logs, but $\frac{1}{x(300-x)}$ is not in that shape.
Partial fractions (the Ch7 method, cover-up or equating coefficients) rewrites it as a sum of those easy pieces, and each one becomes a log.

The chain: decompose → combine logs → exponentiate → untangle

Worked example

Textbook 10.3: $\dfrac{dx}{dt}=\dfrac{x(300-x)}{600}$ , $x=50$ when $t=0$

A population-style model: the variables are almost split already, so all the work is in the $x$ -fraction.

Separate (keep the 600 on the right, simpler): $\dfrac{1}{x(300-x)}\,dx=\dfrac{1}{600}\,dt$ .
Decompose (Ch7): $\dfrac{1}{x(300-x)}=\dfrac{1}{300}\Big(\dfrac{1}{x}+\dfrac{1}{300-x}\Big)$ .
Integrate each side. Note $\int\frac{1}{300-x}\,dx=-\ln(300-x)$ (chain rule sign): $\tfrac{1}{300}\big[\ln x-\ln(300-x)\big]=\tfrac{1}{600}t+C$ . Modulus dropped since $0<x<300$ .
Combine logs with $\ln p-\ln q=\ln\frac{p}{q}$ , and tidy (multiply by 300): $\ln\dfrac{x}{300-x}=\tfrac12 t+C$ .
Find $C$ : at $x=50,\;t=0$ , $C=\ln\frac{50}{250}=\ln 0.2$ .
Exponentiate, then untangle $x$ (cross-multiply, collect $x$ ): $\dfrac{x}{300-x}=0.2\,e^{\frac12 t}$ $\;\Rightarrow\;$ $x\big(1+0.2\,e^{\frac12 t}\big)=60\,e^{\frac12 t}$ .

Answer

x=\dfrac{60\,e^{\frac12 t}}{1+0.2\,e^{\frac12 t}}

Where most students freeze is that last untangle: getting the lone $x$ out of a fraction. Cross-multiply, gather every $x$ term, factor $x$ out, divide.

Common mistake

Three classic slips on this type:

Integrating the fraction before splitting it (there is no clean integral until you split it).
Sign error when combining logs, especially the $-\ln(300-x)$ from the $(300-x)$ piece.
Leaving the answer in $\ln$ form when “ $x$ in terms of $t$ ” was asked. You must exponentiate and get $x$ on its own.

When the question names the technique

Worked example

9709/31 May–Jun 2021 Q10: $\dfrac{dx}{dt}=x^2(1+2x)$ , $x=1$ when $t=0$ ; $t$ in terms of $x$ [11]

The question says “using partial fractions”, so separate with the $x$ -fraction on the left: $\dfrac{1}{x^2(1+2x)}\,dx=dt$ .

Decompose (repeated factor $x^2$ needs both $\frac{A}{x}$ and $\frac{B}{x^2}$ ): $\dfrac{1}{x^2(1+2x)}=\dfrac{1}{x^2}-\dfrac{2}{x}+\dfrac{4}{2x+1}$ .
Integrate: $t=-\dfrac{1}{x}-2\ln x+2\ln(2x+1)+C$ ( $\int\frac{4}{2x+1}=2\ln(2x+1)$ ).
Find $C$ : at $x=1,\;t=0$ , $0=-1-0+2\ln 3+C$ , so $C=1-2\ln 3$ .
Combine logs into one: $2\ln(2x+1)-2\ln x-2\ln 3=2\ln\dfrac{2x+1}{3x}$ .

Answer

t=1-\dfrac{1}{x}+2\ln\!\dfrac{2x+1}{3x}

Checked. Because it asked for $t$ in terms of $x$ , you stop at this $\ln$ form; no exponentiating needed.

Your turn— tap to reveal the worked answer (9709/32 Feb–Mar 2022 Q9 (shorter warm-up))

Answer

y=\dfrac{\sqrt5}{2}\ \text{ at }\ x=3