Antiderivatives

Integration has a power rule, like differentiation, and every table gives it one exception: n = −1. Slide n through that exception and watch where it goes. Then follow one integral a rule at a time, and try your own: type an antiderivative, and the page differentiates it and compares the result with the integrand.

Power rule	$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C, n \neq = - 1$
The exception, n = −1	$\int \frac{d x}{x} = ln ∣ x ∣ + C$

Worked example — Antiderivatives

Constant between the two forms	100.00
Gap to ln x, divided by n + 1	0.9654

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Antiderivatives

An antiderivative of f on an interval is a function F with F′(x) = f(x) at every x of the interval. If F is one, so is F + C for every constant C, and on an interval those are all of them; on a domain in two pieces, such as x ≠ 0 for 1/x, each piece takes its own constant. The indefinite integral ∫f(x) dx names the whole family.

What each symbol means

F_n: the antiderivative of xⁿ through (1, 0): (xⁿ⁺¹ − 1)/(n + 1), which is ∫₁^x tⁿ dt, and ln x at n = −1
C: the constant of integration. Its derivative is 0, so any value gives an antiderivative

Where the formula comes from

Why 1/(n + 1) is the constant to remove. F_n(x) = ∫₁^x tⁿ dt is the area under tⁿ from 1 to x. A small change in n changes every height tⁿ a little, so the area changes a little: F_n cannot jump as n passes −1, and at n = −1 it has to be the area under 1/t from 1 to x, which is ln x. The textbook form is F_n + 1/(n + 1), and that constant is what blows up.
$F_{n} (x) = \int_{1}^{x} t^{n} d t = \frac{x ^{n + 1} - 1}{n + 1} = \frac{x ^{n + 1}}{n + 1} - \to \pm \infty \frac{1}{n + 1}$
The limit, computed. Put h = n + 1 and expand xʰ as a series in h. The gap to ln x is about h(ln x)²/2, which at x = 4 is 0.9609h: the number the last column of the table approaches.
$x^{h} = e^{h l n x} = 1 + h ln x + \frac{h ^{2} ( ln x ) ^{2}}{2} + \dots \Rightarrow F_{n} (x) = \frac{x ^{h} - 1}{h} = ln x + \frac{h ( ln x ) ^{2}}{2} + \dots$

Assumes: The check is numerical. The page differentiates your F exactly, by the rules you would use, then compares F′ with f at 120 points spread across the interval, none of them a whole number, and accepts a relative difference up to 10⁻⁶; a lost factor or a wrong sign is a difference of order 1. Where f is defined and your F is not, the answer fails. Where f is continuous on the whole interval, the page also integrates f numerically between neighbouring points and compares that with the change in your F, which catches an answer that is right piece by piece but jumps.

Problems solved in full

Expand before you integrate 3 steps
Find ∫(x² + 1)² dx.

Open this problem in the tool
1. There is no power rule for a bracket. Multiply it out, so that every term is a power of x:
  
  $(x^{2} + 1)^{2} = x^{4} + 2 x^{2} + 1$
2. Integrate term by term, raising each power by one and dividing by the new power:
  
  $\int (x^{4} + 2 x^{2} + 1) d x = \frac{x ^{5}}{5} + 2 \cdot \frac{x ^{3}}{3} + x + C$
3. Check by differentiating:
  
  $\frac{d}{d x} (\frac{x ^{5}}{5} + \frac{2 x ^{3}}{3} + x) = x^{4} + 2 x^{2} + 1$
Answer

x⁵/5 + 2x³/3 + x + C. Raising the bracket instead, to (x² + 1)³/3, is the classic slip. Its derivative is 2x(x² + 1)², and the 2x comes from inside the bracket, which the power rule never accounts for. Open the problem in the tool and type that answer to watch the two curves part.

$\int (x^{2} + 1)^{2} d x = \frac{x ^{5}}{5} + \frac{2 x ^{3}}{3} + x + C$
Divide before you integrate 3 steps
Find ∫(x³ − 2√x + 5)/x dx for x > 0.

Open this problem in the tool
Show solution
1. Divide each term by x and write it as a power:
  
  $\frac{x ^{3} - 2 x + 5}{x} = x^{2} - 2 x^{- 1/2} + 5 x^{- 1}$
2. The first two are the power rule:
  
  $\int x^{2} d x = \frac{x ^{3}}{3}, - 2 \int x^{- 1/2} d x = - 2 \cdot \frac{x ^{1/2}}{1/2} = - 4 x$
3. The third is x⁻¹, the one power the rule cannot divide by:
  
  $5 \int x^{- 1} d x = 5 ln ∣ x ∣ = 5 ln x (x > 0)$
Answer

x³/3 − 4√x + 5 ln x + C. The bars of ln|x| can go because the problem is set on x > 0, which √x needs anyway.

$\int \frac{x ^{3} - 2 x + 5}{x} d x = \frac{x ^{3}}{3} - 4 x + 5 ln x + C$
Three lines of the table 2 steps
Find ∫(2 cos x − 3 sin x + 4/(1 + x²)) dx.

Open this problem in the tool
Show solution
1. Each term is in the table:
  
  $\int cos x d x = sin x, \int sin x d x = - cos x, \int \frac{d x}{1 + x ^{2}} = arctan x$
2. Keep each constant and replace each function by its antiderivative:
  
  $2 sin x - 3 (- cos x) + 4 arctan x$
Answer

2 sin x + 3 cos x + 4 arctan x + C. The middle term is the one that goes wrong. The antiderivative of sin x is −cos x, so −3 sin x integrates to +3 cos x; the minus comes from differentiating cos x, which gives −sin x.

$\int (2 cos x - 3 sin x + \frac{4}{1 + x ^{2}}) d x = 2 sin x + 3 cos x + 4 arctan x + C$
A condition fixes the constant 2 steps
Find f if f′(x) = 6x² − 4x + 1 and f(1) = 5.

Open this problem in the tool
Show solution
1. Integrate first, keeping C:
  
  $f (x) = \int (6 x^{2} - 4 x + 1) d x = 2 x^{3} - 2 x^{2} + x + C$
2. Then put x = 1 and solve for C:
  
  $f (1) = 2 - 2 + 1 + C = 1 + C = 5 \Rightarrow C = 4$
Answer

f(x) = 2x³ − 2x² + x + 4. The tool checks the condition as well as the derivative. Leave C at 0 and it accepts F′ = f, then reports F(1) = 1, not 5, and tells you to change the constant by 4.

$f (x) = 2 x^{3} - 2 x^{2} + x + 4$
Two integrations, two constants 2 steps
Find f if f″(x) = 12x − 6, f′(0) = 2 and f(0) = −1.

Open this problem in the tool
Show solution
1. Integrate once to get f′, and use the condition on f′ to fix its constant:
  
  $f^{'} (x) = \int (12 x - 6) d x = 6 x^{2} - 6 x + C_{1}, f^{'} (0) = C_{1} = 2$
2. Integrate again, and use the condition on f:
  
  $f (x) = \int (6 x^{2} - 6 x + 2) d x = 2 x^{3} - 3 x^{2} + 2 x + C_{2}, f (0) = C_{2} = - 1$
Answer

f(x) = 2x³ − 3x² + 2x − 1. Each integration brings its own constant, and each condition fixes one of them. The tool opens on the second integration, with f′ as the integrand and f(0) = −1 as the condition.

$f (x) = 2 x^{3} - 3 x^{2} + 2 x - 1$
ln|x| on the negative side 2 steps
Find f if f′(x) = 2/x for x < 0 and f(−1) = 3.

Open this problem in the tool
Show solution
1. For negative x, ln x has no value, but ln(−x) has one, and its derivative is 1/x:
  
  $x < 0 : \frac{d}{d x} ln (- x) = \frac{- 1}{- x} = \frac{1}{x}$
2. So f(x) = 2 ln|x| + C, and the condition fixes C:
  
  $f (x) = 2 ln ∣ x ∣ + C, f (- 1) = 2 ln 1 + C = C = 3$
Answer

f(x) = 2 ln|x| + 3 for x < 0. The condition says nothing about positive x. 1/x is undefined at 0, so a function equal to 2 ln x + 7 for x > 0 would answer the problem just as well.

$f (x) = 2 ln ∣ x ∣ + 3 (x < 0)$
Find ln 4 with a square-root key 4 steps
A calculator has a square-root key and no ln key. Use the limit on this page to find ln 4 to four decimal places. Press n = −0.99 first: the curve through (1, 0) reads 1.3959 at x = 4, still well above ln 4 = 1.3863.

Open this problem in the tool
Show solution
1. Ten presses of the square-root key take 4 to 4 to the power 1/1024, so h = n + 1 = 1/1024:
  
  $4^{1/1024} = 1.001354720, h = n + 1 = \frac{1}{1024}$
2. The curve through (1, 0) at that h, the page's formula with one subtraction and one multiplication:
  
  $F (h) = \frac{4 ^{h} - 1}{h} = 1.3872332$
3. Its gap to ln 4 is about h(ln 4)²/2, the table's last column times h. That is too big for four decimals:
  
  $\frac{h ( ln 4 ) ^{2}}{2} = \frac{0 . 9609}{1024} = 0.000938$
4. One more square root halves h, and with it the gap. Twice the new reading minus the old one cancels the gap's leading term:
  
  $2 F (\frac{h}{2}) - F (h) = 2 \times 1.3867637 - 1.3872332 = 1.386294$
Answer

ln 4 ≈ 1.3863. The combination 2F(h/2) − F(h) removes the error that is proportional to h and leaves one proportional to h², here about 2×10⁻⁷. The page moves n in steps of 0.01, so h = 1/2048 is far past what its slider can show.

$ln 4 \approx 1.3863$

ln x is the power rule's own limit 🖖

Press n = −0.99. The textbook antiderivative of xⁿ is xⁿ⁺¹/(n + 1), which divides by zero at n = −1, and that is why tables list ∫x⁻¹ dx = ln|x| as a rule of its own. The dashed ghost shows what the formula does on the way there: at x = 4 it reads 101.40, and it grows without bound as n closes in. All of that growth is in the constant. Subtract 1/(n + 1), here 100.00, and you have the antiderivative through (1, 0), the solid curve, at 1.3959 against ln 4 = 1.3863. Press Sweep n from −2 to 0 and watch n pass through −1: the ghost leaves the chart, the solid curve never jumps, and at n = −1 it is ln x. It cannot jump, because it is the area under tⁿ from 1 to x, and a small change in n changes every height tⁿ only a little. Grégoire de Saint-Vincent came at the same fact from the other side in his Opus geometricum of 1647: the area under the hyperbola from a to b depends only on b/a. His pupil Alphonse Antonio de Sarasa saw that this makes the area a logarithm.

ln x answers 1/x on only half the line 🖖

Press ln x for 1/x. The trainer wants an antiderivative of 1/x on −4 ≤ x ≤ 4, and ln x has been typed. Wherever ln x exists its derivative is 1/x, so it looks right, and the check refuses it: at x = −2.75, where 1/x is −0.364, ln x has no value. What works on both sides is ln|x|, because ln(−x) differentiates to −1/(−x) = 1/x. Type it, and the upper panel draws two branches with grey copies that move each branch by a different amount. Those would pass too. Type ln|x| + 3sgn(x), which is ln|x| + 3 right of 0 and ln|x| − 3 left of it, and it passes; the panel draws its difference from ln|x| as two flat pieces, at 3 and −3. 1/x is undefined at 0, so nothing ties the constant on the left to the one on the right. Tables write ln|x| + C with one C, but across the whole domain the general antiderivative has a constant for each side. Where the integrand is continuous across the interval, as x³ is, the check refuses an answer that jumps, even when its derivative matches wherever it exists.

Two answers that look nothing alike can both be right 🖖

Press Two forms of one answer. The integrand is (x + 1)². Raise the power and divide, and you get (x + 1)³/3; expand first, and you get x³/3 + x² + x. Set side by side they look like a right answer and a wrong one, and the trainer accepts both. The row Your F minus ours says why: −1/3 at every point, because (x + 1)³/3 expands to x³/3 + x² + x + 1/3. Press 2 sin x cos x for a sharper case. sin²x, −cos²x and −cos(2x)/2 share no visible term, and all three are right, sitting 1 and ½ apart. So the check never decides by comparing your answer with the page's. It differentiates yours, because differentiation is a procedure with one result, and compares that with the integrand. Finding an antiderivative is a search, and two good searches can stop at different constants.

Learning path

Techniques of integration

References (2)

The definition the page checks against (“F is an antiderivative of f if F′(x) = f(x) for all x in the domain of f”), the power rule with its exception (Theorem 4.15: for n ≠ −1, ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C), and the derivation of d/dx ln|x| = 1/x from ln(−x) on x < 0: G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 4.10, “Antiderivatives”.
Grégoire de Saint-Vincent's Opus geometricum (Antwerp, 1647) showed that the area under a rectangular hyperbola over [a, b] equals the area over [c, d] when a/b = c/d, and his pupil Alphonse Antonio de Sarasa made the connection with logarithms: J. J. O'Connor and E. F. Robertson, “Grégoire de Saint-Vincent”, MacTutor History of Mathematics Archive, University of St Andrews (last updated November 2010).