Integrals

Antiderivatives

Integration runs differentiation backward. An antiderivative of \(f\) is any function whose derivative is \(f\), and the indefinite integral names the whole family of them at once:

\[\int \class{var-fn}{f}(x)\,dx = \class{var-gn}{F}(x) + C \quad \text{where} \quad \class{var-gn}{F}'(x) = \class{var-fn}{f}(x)\]

Reading the notation

\(\int\) the integral sign, an elongated S for "sum"

\(dx\) names the variable of integration; it is not optional always write it

\(+C\) constants differentiate to zero, so every answer has an unknown one

Check any answer differentiate it; you should get the integrand back free verification

Indefinite vs. definite a family of functions vs. a single number

The rules that reverse cleanly

Power rule \(\displaystyle\int x^n dx = \frac{x^{n+1}}{n+1} + C\), for \(n \neq -1\)

The exception \(\displaystyle\int x^{-1}dx = \ln|x| + C\), since \(n+1\) would be zero special case

Constant multiple \(\displaystyle\int c\,f\,dx = c\displaystyle\int f\,dx\); constants pull straight out

Sum and difference integrate term by term

No product or quotient rule \(\int fg \neq \int f \int g\); products need substitution or parts common error

Differentiation is a procedure: follow the rules and an answer always appears. Integration is pattern matching, so the goal of the rest of this page is to force unfamiliar integrals into shapes you already recognize.

Definite integrals

Attaching limits turns the family of antiderivatives into one number, the signed area under the curve between those bounds. The fundamental theorem of calculus says you get it by evaluating any antiderivative at both ends:

\[\int_{\class{var-gn}{a}}^{\class{var-gn}{b}} \class{var-fn}{f}(x)\,dx = \class{var-lim}{F}(\class{var-gn}{b}) - \class{var-lim}{F}(\class{var-gn}{a})\]

Properties

Swapping the limits \(\displaystyle\int_b^a f = -\int_a^b f\)

Equal limits \(\displaystyle\int_a^a f\,dx = 0\), no width means no area

Splitting \(\displaystyle\int_a^c f = \int_a^b f + \int_b^c f\)

Signed area area below the axis counts as negative not the same as total area

No \(+C\) the constants cancel in the subtraction

The other half of the theorem

Accumulation \(\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt = f(x)\)

What it means differentiation and integration undo each other the whole point

Variable upper limit if the top is \(g(x)\), multiply by \(g'(x)\) as usual

Net change \(\displaystyle\int_a^b f'(x)\,dx = f(b) - f(a)\), the integral of a rate is the total change

Every technique below works for definite integrals too. The only extra care is with substitution, where the bounds belong to \(x\) and have to be handled deliberately.

Common forms

These are the reference tables. Each is a derivative rule read right to left, so anything you can differentiate on sight belongs here as well.

Powers, exponentials, logarithms

\(\displaystyle\int x^n dx\) \(\dfrac{x^{n+1}}{n+1} + C\), \(n \neq -1\)

\(\displaystyle\int \frac{1}{x}dx\) \(\ln|x| + C\) absolute value required

\(\displaystyle\int e^x dx\) \(e^x + C\)

\(\displaystyle\int b^x dx\) \(\dfrac{b^x}{\ln b} + C\)

\(\displaystyle\int \ln x\,dx\) \(x\ln x - x + C\) by parts

\(\displaystyle\int k\,dx\) \(kx + C\)

Trigonometric

\(\displaystyle\int \sin x\,dx\) \(-\cos x + C\) sign flips

\(\displaystyle\int \cos x\,dx\) \(\sin x + C\)

\(\displaystyle\int \sec^2 x\,dx\) \(\tan x + C\)

\(\displaystyle\int \sec x \tan x\,dx\) \(\sec x + C\)

\(\displaystyle\int \tan x\,dx\) \(-\ln|\cos x| + C\)

\(\displaystyle\int \sec x\,dx\) \(\ln|\sec x + \tan x| + C\) memorize

Forms giving inverse trig

\(\displaystyle\int \frac{dx}{1 + x^2}\) \(\arctan x + C\)

\(\displaystyle\int \frac{dx}{\sqrt{1 - x^2}}\) \(\arcsin x + C\)

\(\displaystyle\int \frac{dx}{a^2 + x^2}\) \(\dfrac{1}{a}\arctan\dfrac{x}{a} + C\)

\(\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}\) \(\arcsin\dfrac{x}{a} + C\)

Recognizing them a lone constant plus or minus a square in the denominator look for this shape

Linear inside shortcut

The pattern \(\displaystyle\int f(ax + b)\,dx = \dfrac{1}{a}F(ax+b) + C\)

\(\displaystyle\int \cos(5x)\,dx\) \(\dfrac{1}{5}\sin(5x) + C\)

\(\displaystyle\int e^{3x}dx\) \(\dfrac{1}{3}e^{3x} + C\)

\(\displaystyle\int (2x + 1)^4 dx\) \(\dfrac{(2x+1)^5}{10} + C\)

Only for linear insides with \(x^2\) or worse inside, you need a real substitution common error

That shortcut is substitution done in your head, and the \(1/a\) out front is there to cancel the chain rule factor differentiation would produce.

u-substitution

Substitution reverses the chain rule. If the integrand contains an inner function alongside its own derivative, renaming that inner function collapses the whole expression into something basic:

\[\int \class{var-fn}{f}\big(\class{var-gn}{g}(x)\big)\class{var-gn}{g}'(x)\,dx = \int \class{var-fn}{f}(\class{var-gn}{u})\,d\class{var-gn}{u}\]

Procedure

1. Choose \(u\) the inner function, whatever sits inside a power, root, exponent, or trig argument

2. Differentiate \(du = g'(x)\,dx\), then solve for \(dx\) if that helps

3. Substitute everything no \(x\) may survive in the new integral check before integrating

4. Integrate in \(u\) it should now match a form from the table above

5. Substitute back replace \(u\) with \(g(x)\), and keep the \(+C\)

Choosing \(u\)

Best candidate the expression whose derivative also appears, up to a constant multiple

Inside a composition the exponent of \(e\), the argument of \(\ln\) or \(\sin\), the base of a power

A denominator good choice when the numerator is close to its derivative

Constants are fine a stray factor of \(2\) can be balanced by a \(\tfrac{1}{2}\) out front

Variables are not a leftover \(x\) cannot be moved outside the integral common error

For a definite integral, convert the bounds as well: \(x = a\) becomes \(u = g(a)\). Then you evaluate directly in \(u\) and never substitute back.

Example: \(\displaystyle\int x\sqrt{x^2 + 5}\,dx\)

Choose \(u = x^2 + 5\), the inside of the root

Differentiate \(du = 2x\,dx\), so \(x\,dx = \tfrac{1}{2}du\)

Rewrite \(\displaystyle\frac{1}{2}\int u^{1/2}\,du\)

Integrate \(\dfrac{1}{2}\cdot\dfrac{2}{3}u^{3/2} = \dfrac{1}{3}u^{3/2}\)

Back-substitute \(\dfrac{1}{3}(x^2+5)^{3/2} + C\) answer

Integration by parts

Integrating the product rule and rearranging gives the tool for products that substitution cannot touch. It trades one integral for another, and the method only pays off if the new one is easier:

\[\int \class{var-fn}{u}\,d\class{var-gn}{v} = \class{var-fn}{u}\class{var-gn}{v} - \int \class{var-gn}{v}\,d\class{var-fn}{u}\]

Choosing \(u\) with LIATE

The ranking Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential

How to use it whichever type comes first in that list becomes \(u\); the rest is \(dv\)

Why it works \(u\) should get simpler when differentiated, \(dv\) should be easy to integrate

\(\displaystyle\int \ln x\,dx\) take \(u = \ln x\) and \(dv = dx\) the standard trick

Definite version \(\displaystyle\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du\)

When to reach for it

Mixed types a polynomial times a trig or exponential factor

A lone log or inverse trig \(\ln x\) and \(\arctan x\) have no basic antiderivative

Repeat if needed \(x^2 e^x\) takes two passes, dropping the power each time

Cyclic case \(e^x\sin x\) returns to itself; solve for the integral algebraically clever finish

Try substitution first it is faster whenever it applies order of attack

Example: \(\displaystyle\int x e^{3x}\,dx\)

Assign by LIATE algebraic beats exponential, so \(u = x\) and \(dv = e^{3x}dx\)

Build the pieces \(du = dx\), \(v = \tfrac{1}{3}e^{3x}\)

Apply the formula \(\dfrac{x}{3}e^{3x} - \displaystyle\int \dfrac{1}{3}e^{3x}dx\)

Finish \(\dfrac{x}{3}e^{3x} - \dfrac{1}{9}e^{3x} + C\) answer

Verify differentiating with the product rule returns \(xe^{3x}\)

Had the assignment been reversed, the remaining integral would have contained \(x^2\) instead of \(x\), which is worse than where you started. That is the signal to swap the choice and try again.

Key takeaways

Summary

Antiderivative \(\int f\,dx = F + C\) with \(F' = f\); always carry the \(+C\)

Power rule \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) except at \(n = -1\), which gives \(\ln|x|\)

Fundamental theorem \(\int_a^b f\,dx = F(b) - F(a)\), a signed area

u-substitution reverses the chain rule; pick \(u\) so that \(du\) is already present

By parts \(\int u\,dv = uv - \int v\,du\), with \(u\) chosen by LIATE

Strategy simplify, match the table, then substitution, then parts

Checking work differentiate the answer; integration is the only topic that grades itself