Recognize a direct antiderivative
A function whose slope is known
The upper curve is \(f(x) = 2x + \tfrac3x\), for \(x > 0\). An antiderivative of \(f\) is a function \(F\) whose slope at every \(x\) equals the height of \(f\) there.
The short dashes show the slope that \(F\) must have at each \(x\), steeper where \(f\) is taller.
The teal curve is a candidate for \(F\). A dash is orange where the candidate runs along it, and red where the candidate cuts across it.
The first candidate is \(x^2\). Its slope is \(2x\), which leaves out the part \(\tfrac3x\) of \(f(x)\), so the dashes are red. Move the Candidate slider to the right.
The second candidate is \(x^2 + 3\ln x\). The slopes of \(x^2\) and \(\ln x\) are \(2x\) and \(\tfrac1x\), so the candidate has slope \(2x + \tfrac3x\) and every dash is orange.
A table read backward
Press each button. It shows a function \(f\) above a function \(F\) whose slope equals \(f\), so in every case \(F' = f\).
If an integrand matches one of these rows of the table of antiderivatives, that \(F\) is an antiderivative, and no other method is needed.
Move the Constant C slider. The teal curve moves up and down, and the dashes stay orange, because adding a constant does not change a slope. Every \(F + C\) is an antiderivative.
Notation
The indefinite integral of \(f\) is every function \(\cb{F}\) with \(F' = f\). Adding a constant does not change a slope, so they are all written \(F + C\).
\[\int f(x)\,dx = \cb{F(x)} + C\]For a power, you add 1 to the exponent and divide by the new exponent \(n + 1\). This row needs \(n \ne -1\), because \(n + 1\) is a divisor.
\[\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\]The power \(x^{-1} = \tfrac1x\) has its own row, a logarithm, where the absolute value lets \(x\) be negative. The function \(e^x\) is its own antiderivative.
\[\begin{gathered} \int \frac{1}{x}\,dx = \ln|x| + C \\ \int e^x\,dx = e^x + C \end{gathered}\]Each trigonometric derivative gives a row when you read it backward.
\[\begin{gathered} \int \cos x\,dx = \sin x + C \\ \int \sin x\,dx = -\cos x + C \\ \int \sec^2 x\,dx = \tan x + C \end{gathered}\]Two rows contain a positive number \(a\), which you read from the integrand. The graph shows the case \(a = 1\).
\[\begin{gathered} \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan\frac{x}{a} + C \\ \int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C \end{gathered}\]A constant factor stays outside the integral, and a sum is integrated term by term. To check an answer, differentiate it: the result must be the integrand.
When it fails
Press Show the power row. Then move the Exponent n slider toward \(-1\). The candidate rises out of the picture, because the divisor \(n + 1\) is close to 0.
Press Set n = −1. The divisor is 0 and the row gives no function. The integrand \(\tfrac1x\) has its own row, \(\ln|x|\), which the power row cannot replace.
Example 1: terms that match rows
Find \(\displaystyle\int \left(8x^3 - 2\sin x\right)dx\).
The faint dashed curves are the two terms. You integrate each one, with its constant factor outside, and add the results.
\[8\int x^3\,dx - 2\int \sin x\,dx\]The power row with \(n = 3\) gives \(\tfrac{x^4}{4}\). The factor 8 stays, so the first term is \(2x^4\).
\[8 \cdot \frac{x^4}{4} = \cb{2x^4}\]The sine row gives \(-\cos x\). The factor \(-2\) turns it into \(+2\cos x\).
\[-2 \cdot (-\cos x) = \cb{2\cos x}\]Add the two terms and the constant. At \(x = 1\), the slope of the answer is \(8 - 2\sin 1\), the height of the curve, and the dashes are orange at every \(x\).
\[\begin{aligned} &\int \left(8x^3 - 2\sin x\right)dx \\ &\quad = \cb{2x^4 + 2\cos x} + C \end{aligned}\]
Example 2: a fraction that is not a logarithm
Find \(\displaystyle\int \frac{6}{x^2 + 4}\,dx\).
At \(x = 0\), the curve is \(\tfrac64 = \tfrac32\) tall, so an antiderivative must have slope \(\tfrac32\) there.
A tempting wrong move: A fraction in \(x\) suggests a logarithm, so take \(6\ln(x^2 + 4)\).
\[\begin{aligned} &\int \frac{6}{x^2+4}\,dx \\ &\quad = 6\ln(x^2 + 4) + C \end{aligned}\]The slope of this candidate is \(\frac{12x}{x^2+4}\), which is 0 at \(x = 0\), so its red tangent there is flat. The red curve shows this slope and leaves the chalk curve, so the dashes are red.
The denominator has the form \(x^2 + a^2\) with \(a^2 = 4\) and \(a = 2\), so the integrand matches the arctangent row. The factor 6 stays outside, and the row has the factor \(\tfrac1a = \tfrac12\).
\[\begin{aligned} 6 \cdot \frac{1}{2}\arctan\frac{x}{2} + C \\ = \cb{3\arctan\tfrac{x}{2}} + C \end{aligned}\]The slope of the answer is \(\tfrac32 \big/ \left(1 + \tfrac{x^2}{4}\right)\), which equals \(\frac{6}{x^2+4}\), so the dashes are orange.
Practice
Find \(\displaystyle\int \left(\frac{4}{\sqrt{1 - x^2}} + 3\sec^2 x\right)dx\), for \(-1 < x < 1\).
Match each term to a row of the table. Then drag x. The orange height is the slope your answer must have there.
Solution
The first term matches the arcsine row with \(a = 1\), and the factor 4 stays outside.
\[\int \frac{4}{\sqrt{1 - x^2}}\,dx = 4\arcsin x\]The second term matches the \(\sec^2 x\) row, and the factor 3 stays outside.
\[\int 3\sec^2 x\,dx = 3\tan x\]Add the two parts and the constant. The slope of a sum is the sum of the slopes, so the answer has slope \(f(x)\) and the dashes are orange.
\[\cb{4\arcsin x + 3\tan x} + C\]
Questions
The graph shows \(f(x) = \cos x\) with its orange height at \(x = 0\). After you answer, a teal candidate appears below with dashes that show the slope an antiderivative must have.
At \(x = 0\) the curve is 1 tall, so an antiderivative has slope 1 there. Which candidate is an antiderivative of \(\cos x\)?
- \(\sin x\)
- \(\cos x\)
- \(-\sin x\)
Answer
\(\sin x\)
The curve \(\sin x + 2\) is the curve \(\sin x\) moved up by 2. Is it also an antiderivative of \(\cos x\)?
- Yes, because moving a curve up does not change its slope
- No, because an integrand has only one antiderivative
- No, because it does not pass through the origin
Answer
Yes, because moving a curve up does not change its slope
Which function is \(\displaystyle\int \frac{8}{16 + x^2}\,dx\)?
- \(2\arctan\dfrac{x}{4} + C\)
- \(8\arctan\dfrac{x}{4} + C\)
- \(8\ln(16 + x^2) + C\)
Answer
\(2\arctan\dfrac{x}{4} + C\)
Summary
You compare the integrand with the rows of the table and write the matching function plus a constant.
This works because each row is a derivative read backward, so the answer has the integrand as its slope at every point.
Use it before any other method, and differentiate the answer to confirm the match.
This is one of five skills in topic 6.14 of Which & Why: AP Calculus.