There is no direct rule for $\displaystyle\int\frac{5x+1}{(x+1)(x-1)}\,dx$. But each partial fraction such as $\dfrac{3}{x-1}$ integrates straight to a logarithm. So the plan is: split first (see Algebraic and Partial Fractions), then integrate each piece.
| Piece | Integral |
|---|---|
| $\dfrac{A}{x - a}$ | $A\ln|x - a| + c$ |
| $\dfrac{A}{bx + c}$ | $\dfrac{A}{b}\ln|bx + c| + c$ |
| $\dfrac{A}{(x - a)^2}$ | $-\dfrac{A}{x - a} + c$ (a power, not a log) |
Find $\displaystyle\int\frac{5x+1}{(x+1)(x-1)}\,dx$.
Show that $\displaystyle\int_2^4\frac{1}{x^2 - 1}\,dx = \tfrac{1}{2}\ln\tfrac{9}{5}$.
Find $\displaystyle\int\frac{x}{(x+1)(2x+1)}\,dx$.
Find $\displaystyle\int\frac{x+5}{(x-1)(x+1)^2}\,dx$, and hence evaluate it between $x = 2$ and $x = 3$.
Show that $\displaystyle\int_2^3\frac{x^2+1}{x^2-1}\,dx = 1 + \ln\tfrac{3}{2}$.
Evaluate $\displaystyle\int_2^3\frac{x^2 + x}{x^2 - 1}\,dx$.
Step 1
Proper? If not, divide first.
Step 2
Factorise the denominator; cancel if possible.
Step 3
Split into partial fractions; check by recombining.
A/(x − a)
$A\ln|x-a|$
A/(bx + c)
$\tfrac{A}{b}\ln|bx+c|$ — divide by $b$.
A/(x − a)²
$-\dfrac{A}{x-a}$ — no log.
Single log
Combine with $\ln a - \ln b = \ln\tfrac{a}{b}$.
Limits
Never across a value that makes a denominator zero.
Find $\displaystyle\int\frac{3}{(x+1)(x-2)}\,dx$.
▶ Show solution
$3 \equiv A(x+1) + B(x-2)$ for $\dfrac{A}{x-2} + \dfrac{B}{x+1}$: $x = 2$: $A = 1$; $x = -1$: $B = -1$.
$\displaystyle\int\left(\frac{1}{x-2} - \frac{1}{x+1}\right)dx = \ln|x-2| - \ln|x+1| + c = \ln\left|\frac{x-2}{x+1}\right| + c$
Find $\displaystyle\int\frac{7x-1}{(x-1)(x+2)}\,dx$.
▶ Show solution
$7x - 1 \equiv A(x+2) + B(x-1)$. $x = 1$: $6 = 3A$, $A = 2$. $x = -2$: $-15 = -3B$, $B = 5$.
$= 2\ln|x-1| + 5\ln|x+2| + c$
Show that $\displaystyle\int_3^5\frac{2}{x^2-4}\,dx = \tfrac{1}{2}\ln\tfrac{15}{7}$.
▶ Show solution
$\dfrac{2}{(x-2)(x+2)} \equiv \dfrac{1/2}{x-2} - \dfrac{1/2}{x+2}$
$\tfrac{1}{2}\Big[\ln(x-2) - \ln(x+2)\Big]_3^5 = \tfrac{1}{2}\left[\ln\tfrac{3}{7} - \ln\tfrac{1}{5}\right] = \tfrac{1}{2}\ln\tfrac{15}{7}$ ✓
Find the exact value of $\displaystyle\int_0^1\frac{x+3}{(x+1)(x+2)}\,dx$, giving your answer as a single logarithm.
▶ Show solution
$x + 3 \equiv A(x+2) + B(x+1)$. $x = -1$: $A = 2$. $x = -2$: $1 = -B$, $B = -1$.
$\Big[2\ln(x+1) - \ln(x+2)\Big]_0^1 = (2\ln 2 - \ln 3) - (0 - \ln 2) = 3\ln 2 - \ln 3 = \ln\tfrac{8}{3}$
Show that $\displaystyle\int_1^2\frac{x+4}{(2x-1)(x+1)}\,dx = \ln\left(2\sqrt{3}\right)$.
▶ Show solution
$x + 4 \equiv A(x+1) + B(2x-1)$. $x = \tfrac{1}{2}$: $\tfrac{9}{2} = \tfrac{3}{2}A$, $A = 3$. $x = -1$: $3 = -3B$, $B = -1$.
$\displaystyle\int\left(\frac{3}{2x-1} - \frac{1}{x+1}\right)dx = \tfrac{3}{2}\ln|2x-1| - \ln|x+1|$
$\left(\tfrac{3}{2}\ln 3 - \ln 3\right) - \left(0 - \ln 2\right) = \tfrac{1}{2}\ln 3 + \ln 2 = \ln\left(2\sqrt{3}\right)$ ✓
Find the exact value of $\displaystyle\int_0^1\frac{2x+3}{(x+1)^2}\,dx$.
▶ Show solution
$2x + 3 \equiv A(x+1) + B$, so $A = 2$ and (at $x = -1$) $B = 1$: $\dfrac{2}{x+1} + \dfrac{1}{(x+1)^2}$.
$\left[2\ln(x+1) - \dfrac{1}{x+1}\right]_0^1 = \left(2\ln 2 - \tfrac{1}{2}\right) - (0 - 1) = 2\ln 2 + \tfrac{1}{2}$
Express $\dfrac{1}{x(x+1)^2}$ in partial fractions, and hence show that $\displaystyle\int_1^2\frac{1}{x(x+1)^2}\,dx = \ln\tfrac{4}{3} - \tfrac{1}{6}$.
▶ Show solution
$1 \equiv A(x+1)^2 + Bx(x+1) + Cx$. $x = 0$: $A = 1$. $x = -1$: $1 = -C$, $C = -1$. $x^2$: $0 = A + B$, $B = -1$.
$\dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{(x+1)^2}$
Integral: $\ln x - \ln(x+1) + \dfrac{1}{x+1}$.
$\left(\ln 2 - \ln 3 + \tfrac{1}{3}\right) - \left(0 - \ln 2 + \tfrac{1}{2}\right) = \ln\tfrac{4}{3} - \tfrac{1}{6}$ ✓
Find the exact value of $\displaystyle\int_3^4\frac{5}{x^2 + x - 6}\,dx$.
▶ Show solution
$x^2 + x - 6 = (x+3)(x-2)$. $5 \equiv A(x-2) + B(x+3)$: $x = 2$: $B = 1$; $x = -3$: $A = -1$.
$\Big[\ln(x-2) - \ln(x+3)\Big]_3^4 = \ln\tfrac{2}{7} - \ln\tfrac{1}{6} = \ln\tfrac{12}{7}$
Find $\displaystyle\int\frac{x^2+1}{x^2-1}\,dx$ and explain why $\displaystyle\int_0^2\frac{x^2+1}{x^2-1}\,dx$ cannot be found this way.
▶ Show solution
As in Worked Example 5, the integrand is $1 + \dfrac{1}{x-1} - \dfrac{1}{x+1}$.
$\displaystyle\int = x + \ln|x-1| - \ln|x+1| + c = x + \ln\left|\frac{x-1}{x+1}\right| + c$
The range $0 \leqslant x \leqslant 2$ contains $x = 1$, where the denominator is zero and the integrand has a vertical asymptote. The area is not finite there, so the integral is not defined.
$\mathrm{f}(x) = \dfrac{2x^2 + 3x - 1}{(x+1)(x-1)}$, $x > 1$.
(a) Show that $\mathrm{f}(x)$ can be written as $A + \dfrac{B}{x-1} + \dfrac{C}{x+1}$ and find $A$, $B$ and $C$. (b) The region $R$ is bounded by the curve $y = \mathrm{f}(x)$, the $x$-axis and the lines $x = 2$ and $x = 4$. Show that the area of $R$ is $4 + \ln 15$. (c) Explain why the condition $x > 1$ is needed.
▶ Show solution
(a) $(x+1)(x-1) = x^2 - 1$. Then $2x^2 + 3x - 1 = 2(x^2 - 1) + 3x + 1$, so $\mathrm{f}(x) = 2 + \dfrac{3x+1}{x^2-1}$.
$3x + 1 \equiv B(x+1) + C(x-1)$. $x = 1$: $4 = 2B$, $B = 2$. $x = -1$: $-2 = -2C$, $C = 1$.
$\mathrm{f}(x) = 2 + \dfrac{2}{x-1} + \dfrac{1}{x+1}$, so $A = 2$, $B = 2$, $C = 1$.
(b) For $x > 1$ every term is positive, so the curve is above the axis.
$\Big[2x + 2\ln(x-1) + \ln(x+1)\Big]_2^4 = (8 + 2\ln 3 + \ln 5) - (4 + 0 + \ln 3)$
$= 4 + \ln 3 + \ln 5 = 4 + \ln 15$ ✓
(c) $\mathrm{f}$ is undefined at $x = \pm 1$. Restricting to $x > 1$ keeps the curve on one continuous branch, so the region between $x = 2$ and $x = 4$ is well defined and the modulus signs in the logs can be dropped.