Integrating Using Partial Fractions
Repeated and irreducible-quadratic factors each demand a different partial-fraction shape — and a different integration technique once you get there.
Repeated linear factors
From Pure Year 2, a repeated linear factor $(ax+b)^2$ in a denominator contributes two terms to the partial fraction decomposition: one with denominator $(ax+b)$ and one with denominator $(ax+b)^2$. This matters here because each piece integrates completely differently.
Write $\dfrac{5x-1}{(x-1)^2(x+2)}=\dfrac A{x-1}+\dfrac B{(x-1)^2}+\dfrac C{x+2}$. Multiplying through: $5x-1=A(x-1)(x+2)+B(x+2)+C(x-1)^2$.
Setting $x=1$: $4=3B\Rightarrow B=\dfrac43$. Setting $x=-2$: $-11=9C\Rightarrow C=-\dfrac{11}9$. Comparing $x^2$ coefficients: $0=A+C\Rightarrow A=\dfrac{11}9$.
So $\dfrac{5x-1}{(x-1)^2(x+2)}=\dfrac{11/9}{x-1}+\dfrac{4/3}{(x-1)^2}-\dfrac{11/9}{x+2}$.
Integrating term by term (the middle term uses the reverse power rule, not a logarithm, since it's a power of $(x-1)$ rather than $\frac1{x-1}$):
$$\int\frac{5x-1}{(x-1)^2(x+2)}\,dx=\frac{11}9\ln|x-1|-\frac43(x-1)^{-1}-\frac{11}9\ln|x+2|+c$$
The term $\dfrac{B}{(x-1)^2}=B(x-1)^{-2}$ integrates to $-B(x-1)^{-1}$ (reverse power rule), not $B\ln|x-1|$ — that mistake only applies to a $\dfrac1{x-1}$ term (power $-1$), never to a $\dfrac1{(x-1)^2}$ term (power $-2$).
Irreducible quadratic factors
If the denominator contains a quadratic factor with no real roots (irreducible), give it a numerator of the form $Bx+C$ (not just a constant) in the partial fraction decomposition. When integrating that piece, split it into two parts: one proportional to $\dfrac{2x+(\text{linear coeff})}{\text{quadratic}}$, which integrates to a logarithm via the reverse chain rule, and a remaining constant-numerator piece, which integrates to an $\arctan$ using the results from the previous lesson.
Write $\dfrac{6x^2+2}{(x-1)(x^2+4)}=\dfrac A{x-1}+\dfrac{Bx+C}{x^2+4}$. Multiplying through: $6x^2+2=A(x^2+4)+(Bx+C)(x-1)$.
Setting $x=1$: $8=5A\Rightarrow A=\dfrac85$. Comparing $x^2$ coefficients: $6=A+B\Rightarrow B=\dfrac{22}5$. Comparing constants: $2=4A-C\Rightarrow C=4A-2=\dfrac{22}5$.
So $\dfrac{6x^2+2}{(x-1)(x^2+4)}=\dfrac{8/5}{x-1}+\dfrac{\frac{22}5x+\frac{22}5}{x^2+4}$.
Split the second term: $\dfrac{22}5\cdot\dfrac x{x^2+4}+\dfrac{22}5\cdot\dfrac1{x^2+4}$. The first piece is $\dfrac{11}5\cdot\dfrac{2x}{x^2+4}$, which integrates (reverse chain rule) to $\dfrac{11}5\ln(x^2+4)$; the second integrates (with $a=2$) to $\dfrac{22}5\times\dfrac12\arctan\left(\dfrac x2\right)=\dfrac{11}5\arctan\left(\dfrac x2\right)$.
$$\int\frac{6x^2+2}{(x-1)(x^2+4)}\,dx=\frac85\ln|x-1|+\frac{11}5\ln(x^2+4)+\frac{11}5\arctan\left(\frac x2\right)+c$$
Improper algebraic fractions
If the numerator's degree is greater than or equal to the denominator's degree, perform polynomial division (or an equivalent algebraic split) first, so the fraction becomes a polynomial plus a proper fraction — only the proper fraction needs partial fractions; the polynomial part integrates directly using the power rule.
The numerator and denominator both have degree $2$, so this is improper. Write $\dfrac{x^2+3x-1}{x(x+1)}=\dfrac{x^2+3x-1}{x^2+x}$, and note $x^2+3x-1=(x^2+x)+2x-1$, so the fraction $=1+\dfrac{2x-1}{x(x+1)}$.
Partial fractions for the remainder: $\dfrac{2x-1}{x(x+1)}=\dfrac Ax+\dfrac B{x+1}$. Multiplying through: $2x-1=A(x+1)+Bx$. Setting $x=0$: $A=-1$. Setting $x=-1$: $-3=-B\Rightarrow B=3$.
So the integrand is $1-\dfrac1x+\dfrac3{x+1}$, giving $\displaystyle\int\left(1-\frac1x+\frac3{x+1}\right)dx=x-\ln|x|+3\ln|x+1|+c$.
Exam-style questions
Question 1 8 marks
Express $\dfrac{4x^2-3x-4}{(x+1)^2(x-2)}$ in partial fractions, and hence find $\displaystyle\int\frac{4x^2-3x-4}{(x+1)^2(x-2)}\,dx$.
$\dfrac{4x^2-3x-4}{(x+1)^2(x-2)}=\dfrac A{x+1}+\dfrac B{(x+1)^2}+\dfrac C{x-2}$; $4x^2-3x-4=A(x+1)(x-2)+B(x-2)+C(x+1)^2$. (M1, A1)
$x=-1$: $3=-3B\Rightarrow B=-1$. $x=2$: $6=9C\Rightarrow C=\dfrac23$. Comparing $x^2$: $4=A+C\Rightarrow A=\dfrac{10}3$. (M1, A1, A1)
$\displaystyle\int\left(\frac{10/3}{x+1}-\frac1{(x+1)^2}+\frac{2/3}{x-2}\right)dx=\frac{10}3\ln|x+1|+(x+1)^{-1}+\frac23\ln|x-2|+c$. (M1, A1, A1)
Question 2 7 marks
Express $\dfrac{3x+5}{(x+2)(x^2+1)}$ in partial fractions, and hence find $\displaystyle\int\frac{3x+5}{(x+2)(x^2+1)}\,dx$.
$\dfrac{3x+5}{(x+2)(x^2+1)}=\dfrac A{x+2}+\dfrac{Bx+C}{x^2+1}$; $3x+5=A(x^2+1)+(Bx+C)(x+2)$. (M1, A1)
$x=-2$: $-1=5A\Rightarrow A=-\dfrac15$. Comparing $x^2$: $0=A+B\Rightarrow B=\dfrac15$. Constants: $5=A+2C\Rightarrow C=\dfrac{13}5$. (M1, A1, A1)
$\displaystyle\int\left(\frac{-1/5}{x+2}+\frac{\frac15x+\frac{13}5}{x^2+1}\right)dx=-\frac15\ln|x+2|+\frac1{10}\ln(x^2+1)+\frac{13}5\arctan x+c$. (M1, A1)