Integrate the decomposed expression
Each piece has a logarithm
The chalk curve is \(\dfrac{5x - 7}{(x - 3)(x + 1)}\), the sum of the faint pieces \(\dfrac{2}{x - 3}\) and \(\dfrac{3}{x + 1}\). The teal region under the curve, from \(x = 4\) to the upper limit, is the integral to find.
Integrate each piece on its own: the integral of \(\dfrac{2}{x - 3}\) is \(2\ln|x - 3|\), and the integral of \(\dfrac{3}{x + 1}\) is \(3\ln|x + 1|\). The lower curve is the sum of these two logarithms, and at each x its slope is the height of the chalk curve.
Drag the orange handle. The orange segment shows how far the lower curve rises from \(x = 4\) to the upper limit, and this rise equals the teal area. From \(x = 4\) to \(x = 6\), the area is \(2\ln 3 + 3\ln 7 - 3\ln 5\). As a check, this is about 3.21.
A sum of simple antiderivatives
The integral of a sum is the sum of the integrals, so you can integrate a partial-fraction form one piece at a time. A constant over a linear factor has a logarithm for its antiderivative, and the lower curve is the sum of these logarithms.
Drag the orange handle. The short orange line shows the slope of the lower curve, which is the height of the chalk curve at the same x.
Each logarithm has a vertical asymptote at its own dashed line, where its factor is 0.
Notation
Integrate each term. A constant over a linear factor gives that constant times a logarithm.
\[\int \frac{\cy{A}}{x - a}\,dx = \cy{A}\ln|x - a| + C\]Keep the absolute value. The factor \(x - a\) is negative on one side of \(a\), and the logarithm of a negative number does not exist. The derivative of \(\ln|x - a|\) is \(\dfrac{1}{x - a}\) on both sides.
\[\frac{d}{dx}\ln|x - a| = \frac{1}{x - a}\]If the factor is \(px + q\), divide the constant by \(p\), because the inside function \(px + q\) has the derivative \(p\).
\[\int \frac{\cy{A}}{px + q}\,dx = \frac{\cy{A}}{p}\ln|px + q| + C\]Add the antiderivatives of all the terms, and write one constant at the end. If division gave a polynomial part, integrate it with the power rule.
For a definite integral, subtract the value of the sum at the lower limit \(c\) from its value at the upper limit \(d\). Both limits must lie on the same side of every dashed line, because the function is not defined there.
\[\int_c^d r(x)\,dx = \cb{F}(d) - \cb{F}(c)\]
When it fails
Move the lower limit toward the dashed line at \(x = 3\). The teal region grows taller without bound, and the lower curve falls toward \(-\infty\).
The curve is not defined at \(x = 3\), so the area beside the dashed line has no finite value. The subtraction \(F(d) - F(c)\) measures an area only when both limits lie on one side of the dashed line.
Example 1: Absolute values on an interval
Find \(\displaystyle\int \frac{-3x - 1}{(x - 1)(x - 3)}\,dx\) for \(1 < x < 3\). The pieces are \(\dfrac{2}{x - 1}\) and \(\dfrac{-5}{x - 3}\).
Test the pieces. Over the common denominator, their sum has the original numerator.
\[2(x - 3) - 5(x - 1) = -3x - 1\]Integrate each piece.
\[\begin{gathered} \int \frac{2}{x - 1}\,dx = 2\ln|x - 1| \\ \int \frac{-5}{x - 3}\,dx = -5\ln|x - 3| \end{gathered}\]Add the antiderivatives, and write one constant.
\[\begin{aligned} &\int \frac{-3x - 1}{(x - 1)(x - 3)}\,dx \\ &\quad = 2\ln|x - 1| - 5\ln|x - 3| + C \end{aligned}\]On this interval, \(x - 3\) is negative, so \(\ln(x - 3)\) does not exist. The bars make \(|x - 3|\) equal to \(3 - x\), which is positive. The red curve, the formula without the bars, has no part on the interval. It appears only to the right of \(x = 3\), on top of \(F\).
Example 2: A factor with a coefficient
Find \(\displaystyle\int_0^1 \frac{14x + 17}{(2x + 1)(x + 3)}\,dx\). The pieces are \(\dfrac{4}{2x + 1}\) and \(\dfrac{5}{x + 3}\).
Test the pieces. Over the common denominator, their sum has the original numerator.
\[4(x + 3) + 5(2x + 1) = 14x + 17\]A tempting wrong move: Integrate the first piece as a logarithm of the factor, with the constant 4 in front.
\[\begin{aligned} &\int_0^1 \frac{4}{2x + 1}\,dx \\ &\quad = \Big[4\ln|2x + 1|\Big]_0^1 \\ &\quad = 4\ln 3 \approx 4.39 \end{aligned}\]The derivative of \(4\ln|2x + 1|\) is \(\dfrac{8}{2x + 1}\), the red curve, which is twice as tall as the piece. The piece is at most 4 tall on an interval of width 1, so its area is less than 4. The wrong value \(4\ln 3\) is more than 4.
The inside function \(2x + 1\) has the derivative 2, so divide the constant by 2.
\[\begin{aligned} &\int_0^1 \frac{4}{2x + 1}\,dx = \Big[2\ln|2x + 1|\Big]_0^1 \\ &\quad = 2\ln 3 \approx 2.20 \end{aligned}\]Integrate the second piece between the same limits.
\[\begin{aligned} &\int_0^1 \frac{5}{x + 3}\,dx = \Big[5\ln|x + 3|\Big]_0^1 \\ &\quad = 5\ln 4 - 5\ln 3 \end{aligned}\]Add the two values.
\[\begin{aligned} &2\ln 3 + 5\ln 4 - 5\ln 3 \\ &\quad = 5\ln 4 - 3\ln 3 \approx 3.64 \end{aligned}\]
Practice
Find \(\displaystyle\int_3^6 \frac{4x + 1}{(x - 2)(x + 1)}\,dx\). The pieces are \(\dfrac{3}{x - 2}\) and \(\dfrac{1}{x + 1}\).
Integrate each piece with its own logarithm. The teal area is the value of this sum at x = 6 minus its value at x = 3.
Solution
Integrate each piece.
\[\begin{aligned} &\int \left(\frac{3}{x - 2} + \frac{1}{x + 1}\right)dx \\ &\quad = 3\ln|x - 2| + \ln|x + 1| \end{aligned}\]Substitute each limit and subtract. The term \(3\ln 1\) is 0, because \(\ln 1 = 0\).
\[\begin{aligned} &\big(3\ln 4 + \ln 7\big) - \big(3\ln 1 + \ln 4\big) \\ &\quad = 2\ln 4 + \ln 7 \end{aligned}\]Combine the logarithms. As a check, this is about 4.72.
\[2\ln 4 + \ln 7 = \ln 112\]
Questions
The top curve is \(\dfrac{3x - 6}{(x + 2)(x - 1)}\) for \(x > 1\). The lower curve is its antiderivative \(F(x) = 4\ln|x + 2| - \ln|x - 1|\).
On the interval \(1 < x < 2\), the top curve is below the axis. What does \(F\) do on this interval?
- It decreases.
- It increases.
- It stays constant.
Answer
It decreases.
At \(x = 2\), the top curve crosses the axis from below to above. What does \(F\) have at \(x = 2\)?
- A local minimum.
- A local maximum.
- A vertical asymptote.
Answer
A local minimum.
Find \(\displaystyle\int \left(\frac{6}{3x - 1} + \frac{2}{x + 4}\right)dx\).
- \(2\ln|3x - 1| + 2\ln|x + 4| + C\)
- \(6\ln|3x - 1| + 2\ln|x + 4| + C\)
- \(2\ln|3x - 1| + \tfrac23\ln|x + 4| + C\)
Answer
\(2\ln|3x - 1| + 2\ln|x + 4| + C\)
Summary
The teal area is the rise of the lower curve, which is the sum of one logarithm of an absolute value for each linear factor.
This works because the derivative of \(\ln|px + q|\) is \(\dfrac{p}{px + q}\), so each logarithm, divided by \(p\), differentiates back to its piece.
Use it to finish an integral of a rational function after you split it.
This is one of four skills in topic 6.12 of Which & Why: AP Calculus.