Indefinite Integrals and Basic Antiderivatives

Running a derivative backward

Think of the car: if $s(t)$ is position and $v(t) = s'(t)$ is velocity, then recovering $s(t)$ from $v(t)$ is exactly this problem.

The graph below shows $F(x) = x^2$ and $f(x) = 2x$. Notice that $f$ is the slope of $F$ at every point: where $F$ flattens out at $x = 0$, $f$ crosses zero.

The best way to check an antiderivative is to differentiate it. You do not need to integrate anything to verify an answer.

The constant of integration

Both $x^2$ and $x^2 + 5$ have derivative $2x$. So $2x$ has more than one antiderivative, in fact infinitely many.

Slide $C$ below. Every curve has the same slope at a given $x$, because each one is an antiderivative of $f(x) = x$.

To pick out one antiderivative from the family you need one extra piece of information, usually a point on its graph. This is called an initial condition, and the function you get is a particular antiderivative.

The standard antiderivative table

Every derivative rule you know gives an antiderivative rule when read backward. There is nothing new to memorize, just the derivative table in reverse.

The absolute value in $\ln|x|$ matters. For $x<0$, $\dfrac{d}{dx}\ln(-x) = \dfrac{1}{x}$ too, so $\ln|x|$ is an antiderivative of $\dfrac{1}{x}$ on both $(-\infty, 0)$ and $(0, \infty)$.

Radicals and fractions: before using the power rule, rewrite everything as a power of $x$: $\sqrt{x} = x^{1/2}$, $\sqrt[3]{x^2} = x^{2/3}$, $\dfrac{1}{x^4} = x^{-4}$, $\dfrac{1}{\sqrt{x}} = x^{-1/2}$.

Linearity: sums and constant multiples

The graph shows $f(x) = 6x^2 - 5\cos x + 2$ and its antiderivative $F(x) = 2x^3 - 5\sin x + 2x$, which is the next example.

Splitting fractions. You may split a numerator over a single-term denominator, because $\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c}$.

A linear function inside: the $\tfrac{1}{a}$ rule

The table handles $e^x$, but what about $e^{2x}$? Try $e^{2x}$ itself as the answer. By the chain rule, $(e^{2x})' = 2e^{2x}$, which is twice too big. So divide by $2$.

For example, $\displaystyle\int \cos(5x)\,dx = \frac{\sin(5x)}{5} + C$, $\ \displaystyle\int e^{3x-1}dx = \frac{e^{3x-1}}{3} + C$, and $\ \displaystyle\int \frac{1}{2x+7}\,dx = \frac{1}{2}\ln|2x+7| + C$.

Watch the amplitude below: $\sin(kx)$ has antiderivative $-\dfrac{\cos(kx)}{k}$, so when $k$ grows the antiderivative gets flatter.

For a base other than $e$, first write $a^{bx} = (a^b)^x$ or just use the rule with a linear inside: $\displaystyle\int 5^{2x}dx = \frac{5^{2x}}{2\ln 5} + C$.

The reverse chain rule

What if the inside function is not linear? The chain rule says $\dfrac{d}{dx}F(g(x)) = f(g(x))\cdot g'(x)$. Read backward:

Here is the procedure:

1. Look for an inside function $u = g(x)$ whose derivative (up to a constant factor) also appears in the integrand.

2. Rewrite the integral in terms of $u$ and $du$, adjusting by a constant factor if needed.

3. Integrate in $u$.

4. Substitute $u = g(x)$ back.

5. Check by differentiating.

The graph shows $f(x) = 2x\cos(x^2)$ and its antiderivative $F(x) = \sin(x^2)$. The next example uses the same pattern.

Typical patterns: $\displaystyle\int (\ln x)^n\,\frac{dx}{x}$ (with $u = \ln x$), $\displaystyle\int g'(x)\,e^{g(x)}dx$, and $\displaystyle\int g'(x)\,\big(g(x)\big)^n dx$.

Pay attention to the composite structure when functions are nested. In $\displaystyle\int \frac{\ln(\ln x)}{x\ln x}\,dx$, first let $u = \ln x$ (so $du = \dfrac{dx}{x}$), which turns it into $\displaystyle\int \frac{\ln u}{u}\,du$. Then a second substitution $w = \ln u$ finishes it.

The $\dfrac{u'}{u}$ pattern and logarithms

The most common form of the reverse chain rule is $f(u) = \dfrac{1}{u}$.

If the numerator is a constant multiple of the denominator's derivative, adjust with that constant. For example, $\displaystyle\int\frac{x}{x^2+3}\,dx = \frac{1}{2}\ln(x^2+3) + C$, because $(x^2+3)' = 2x$ and $x = \tfrac{1}{2}\cdot 2x$. The same idea gives $\displaystyle\int\frac{4x-6}{x^2-3x+1}\,dx = 2\ln|x^2-3x+1| + C$, since $4x - 6 = 2(2x-3)$. Another classic is $\displaystyle\int \tan x\,dx = \int\frac{\sin x}{\cos x}\,dx = -\ln|\cos x| + C$.

When the numerator is almost the derivative, add and subtract to force it.

Rewrite first: division and identities

Many integrands that look hard become standard after some algebra or trigonometry. Before integrating, ask yourself:

• Can I split the fraction? (single-term denominator)

• Can I divide the polynomials? (degree of top $\ge$ degree of bottom)

• Can I use a trig identity, such as $\sin^2 x + \cos^2 x = 1$ or $1 + \tan^2 x = \sec^2 x$?

Below, both curves are the same function written two ways: $\dfrac{x^2-3x+5}{x-1}$ and $x - 2 + \dfrac{3}{x-1}$. The curves overlap exactly, so the rewriting costs nothing.

Arctangent and arcsine forms: completing the square

The table has $\dfrac{1}{1+x^2}$ and $\dfrac{1}{\sqrt{1-x^2}}$. With a constant $a > 0$ and a variable $u$, the general versions are:

If the denominator is $1 + (kx)^2$, use the linear inside rule from earlier: $\displaystyle\int\frac{3}{1+25x^2}\,dx = \frac{3}{5}\arctan(5x) + C$. If the denominator is a quadratic with an $x$-term, complete the square first.

If the leading coefficient is not $1$, factor it out first. For instance $3x^2 + 6x + 12 = 3\big[(x+1)^2 + 3\big]$, and then $a = \sqrt{3}$.

When there is also an $x$ in the numerator of a radical, shift the variable and split into a power piece and an arcsine piece.

Guess and check with the product rule

There is no product rule for integrals. But you can often guess the shape of an antiderivative of a product, differentiate the guess with the product rule, and adjust the unknown coefficients until it matches.

Absolute values and piecewise integrands. If $f$ is piecewise, such as $|x - 1|$, integrate each piece separately. An antiderivative is differentiable, so it must be continuous, which forces the constants on different pieces to match at the break point.

Going back two derivatives

Sometimes you are given $f''$. Integrate once to get $f'$, then again to get $f$. Each integration brings its own constant, so you need two conditions.

Conditions about the graph translate into equations:

• a point of inflection at $x = x_0$ means $f''(x_0) = 0$ (and $f''$ changes sign there);

• a horizontal tangent at $x_0$ means $f'(x_0) = 0$;

• a horizontal tangent line that passes through $(0, 5)$ is the line $y = 5$, so $f(x_0) = 5$ at the point of tangency.

If $f''$ is rational, think backward from the quotient rule: differentiating a quotient raises the power in the denominator by one, and you can confirm any guess for $f'$ by differentiating it.

Back to the road trip. Say Mateo's phone logs a speed of $v(t) = 2t + 30$ ft/s. Antidifferentiating gives $s(t) = t^2 + 30t + C$. The speed record alone cannot tell him the distance, because $C$ is where the odometer started. One reading, such as the odometer value at $t = 0$, pins down $C$ and gives the exact position function.

Summary