Partial fractions and reduction formulaeAQA A-Level Further Maths: Revision notes

A rational function whose denominator factorises can be split into partial fractions, each easy to integrate.

For $\frac{5x+1}{(x-1)(x+2)}\equiv\frac{A}{x-1}+\frac{B}{x+2}$, multiply through: $5x+1\equiv A(x+2)+B(x-1)$. Put $x=1$: $6=3A$, so $A=2$. Put $x=-2$: $-9=-3B$, so $B=3$.

Then $\int\frac{5x+1}{(x-1)(x+2)}\,dx=2\ln|x-1|+3\ln|x+2|+c$, and $\int_2^3\frac{5x+1}{(x-1)(x+2)}\,dx=\left[2\ln(x-1)+3\ln(x+2)\right]_2^3=3\ln5-4\ln2=\ln\frac{125}{16}.$ Use $\int\frac{k}{ax+b}\,dx=\frac{k}{a}\ln|ax+b|$. Combine logarithms with $\ln p+\ln q=\ln pq$ and $k\ln p=\ln p^k$.

If the denominator contains a factor $x^2+c$ (or $ax^2+c$) that does not factorise, its partial fraction has a linear numerator: $\frac{5x^2+4x+9}{(x+1)(x^2+4)}\equiv\frac{A}{x+1}+\frac{Bx+C}{x^2+4}.$ Multiply through: $5x^2+4x+9\equiv A(x^2+4)+(Bx+C)(x+1)$.

• $x=-1$: $10=5A$, so $A=2$.
• Compare $x^2$: $5=A+B$, so $B=3$.
• Compare constants: $9=4A+C$, so $C=1$.

So $\mathrm{g}(x)=\frac{2}{x+1}+\frac{3x+1}{x^2+4}$. You can use any mix of substituting values and comparing coefficients.

Split $\frac{Bx+C}{x^2+c}$ into two integrals: $\int\frac{Bx}{x^2+c}\,dx=\frac B2\ln(x^2+c),\qquad\int\frac{C}{x^2+c}\,dx=\frac{C}{\sqrt c}\arctan\frac{x}{\sqrt c}.$ (The first is a logarithm because $Bx$ is a multiple of the derivative $2x$ of the denominator; the second is the standard arctan form.) For $ax^2+c$ divide through by $a$ first.

Example: $\int\frac{3x+1}{x^2+4}\,dx=\frac32\ln(x^2+4)+\frac12\arctan\frac x2+c$, so $\int_0^2\mathrm{g}(x)\,dx=2\ln3+\frac32\ln2+\frac\pi8.$

A reduction formula expresses an integral $I_n$ in terms of an integral with a smaller index, such as $I_{n-1}$ or $I_{n-2}$. You derive it by integration by parts (sometimes also a trigonometric identity).

Example: $I_n=\int_0^1x^n\mathrm{e}^x\,dx$. By parts with $u=x^n$ and $\frac{dv}{dx}=\mathrm{e}^x$: $I_n=\left[x^n\mathrm{e}^x\right]_0^1-n\int_0^1x^{n-1}\mathrm{e}^x\,dx=\mathrm{e}-nI_{n-1}\quad(n\ge1).$ The process stops at a base case such as $I_0$, which can be integrated directly.

Work from the base case upwards.

$I_0=\mathrm{e}-1$, so $I_1=\mathrm{e}-I_0=1$, $I_2=\mathrm{e}-2I_1=\mathrm{e}-2$, and $I_3=\mathrm{e}-3I_2=6-2\mathrm{e}$.

Check: $6-2\mathrm{e}\approx0.563$, a sensible area for $x^3\mathrm{e}^x$ on $[0,1]$.

Keep exact forms ($\mathrm{e}$, $\pi$) throughout and write each step so the repeated use of the formula is visible.

For $I_n=\int_0^{\pi/2}\sin^nx\,dx$, write $\sin^nx=\sin^{n-1}x\sin x$ and integrate by parts with $u=\sin^{n-1}x$, $\frac{dv}{dx}=\sin x$: $I_n=\left[-\sin^{n-1}x\cos x\right]_0^{\pi/2}+(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,dx.$ The bracket is zero at both limits. Replacing $\cos^2x$ by $1-\sin^2x$ gives $I_n=(n-1)(I_{n-2}-I_n)$, so $I_n=\frac{n-1}{n}I_{n-2}\qquad(n\ge2).$ With $I_0=\frac\pi2$ and $I_1=1$: $I_4=\frac34\cdot\frac12\cdot\frac\pi2=\frac{3\pi}{16}$, $I_6=\frac{5\pi}{32}$ and $I_5=\frac45\cdot\frac23=\frac{8}{15}$.