Unit 5, lesson 46

Integration Rules for Indefinite Integrals

The basic rules for finding indefinite integrals, from constants and powers to logarithms, exponentials and trigonometric functions, together with the intervals on which each one holds.

  • Theory part 91
  • Practice part 92
  • Indefinite Integrals

Theory, part 91

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Practice, part 92

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Every derivative formula you know, read backward, is an integration formula. Since the derivative of sin⁡x\sin x is cos⁡x\cos x, an antiderivative of cos⁡x\cos x is sin⁡x\sin x. The rules below collect these reversed formulas into a short table, and with the sum and constant multiple rules they handle a large share of everyday integrals. Each rule holds on an interval where the integrand is defined, and that detail matters more than it looks.

Key terms

Integral of a constant

Definition For any constant kk,

∫k dx=kx+Con R.\int k\,dx = kx + C \quad \text{on } \mathbb{R}.

In plain words A constant rate of change gives a straight line.

Example The integral of 77 is 7x+C7x + C, since the derivative of 7x7x is 77. In particular the integral of 00 is just the constant CC.

Constant multiple rule

Definition If FF is an antiderivative of ff on an interval II and cc is a constant, then cFcF is an antiderivative of cfcf on II. For c≠0c \neq 0 this is written

∫cf(x) dx=c∫f(x) dx.\int c f(x)\,dx = c\int f(x)\,dx.

In plain words A constant factor can be moved out in front of the integral sign.

Example On R\mathbb{R}

∫5cos⁡x dx=5∫cos⁡x dx=5sin⁡x+C.\int 5\cos x\,dx = 5\int \cos x\,dx = 5\sin x + C.

Sum and difference rules

Definition If FF and GG are antiderivatives of ff and gg on an interval II, then F+GF + G and F−GF - G are antiderivatives of f+gf + g and f−gf - g on II,

∫(f(x)±g(x)) dx=∫f(x) dx±∫g(x) dx.\int \bigl(f(x) \pm g(x)\bigr)\,dx = \int f(x)\,dx \pm \int g(x)\,dx.

There is no such rule for products or quotients.

In plain words Integrate a sum one term at a time. One constant CC at the end covers all of them.

Example Term by term,

∫(cos⁡x−2x) dx=sin⁡x−x2+C.\int (\cos x - 2x)\,dx = \sin x - x^2 + C.

By contrast, integrating each factor of xcos⁡xx\cos x separately gives the wrong guess 12x2sin⁡x+C\tfrac{1}{2}x^2\sin x + C, whose derivative is not xcos⁡xx\cos x.

Power rule

Definition For a real number n≠−1n \neq -1,

∫xn dx=xn+1n+1+C,\int x^n\,dx = \frac{x^{n+1}}{n + 1} + C,

where the interval depends on nn.

  1. If n=0,1,2,…n = 0, 1, 2, \ldots, the formula holds on R\mathbb{R}.
  2. If n=−2,−3,…n = -2, -3, \ldots, it holds on (0,∞)(0, \infty) and on (−∞,0)(-\infty, 0), separately.
  3. If n=pqn = \frac{p}{q} is a fraction in lowest terms with qq odd and q>1q > 1, then xnx^n is also defined for x<0x < 0 through the odd root xq\sqrt[q]{x}. The formula holds on R\mathbb{R} when n>0n > 0, and on (0,∞)(0, \infty) and on (−∞,0)(-\infty, 0) separately when n<0n < 0.
  4. For every other real nn, that is, a fraction with even denominator in lowest terms or an irrational number, xnx^n is not defined for x<0x < 0 and the formula holds on (0,∞)(0, \infty).

For example

∫x1/3 dx=34x4/3+Con R,∫x dx=23x3/2+Con (0,∞).\int x^{1/3}\,dx = \frac{3}{4}x^{4/3} + C \quad \text{on } \mathbb{R}, \qquad \int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} + C \quad \text{on } (0, \infty).

In plain words Add one to the exponent and divide by the new exponent. It fails only for n=−1n = -1, where you would divide by zero.

Example On (0,∞)(0, \infty),

∫(3x2−4x3+x)dx=x3+2x2+23x3/2+C.\int \left(3x^2 - \frac{4}{x^3} + \sqrt{x}\right)dx = x^3 + \frac{2}{x^2} + \frac{2}{3}x^{3/2} + C.

The middle term comes from

−4⋅x−2−2=2x−2.-4 \cdot \frac{x^{-2}}{-2} = 2x^{-2}.

Differentiating gives back 3x2−4x−3+x1/23x^2 - 4x^{-3} + x^{1/2}.

Logarithmic rule

Definition On any interval not containing 00,

∫1x dx=ln⁡∣x∣+C.\int \frac{1}{x}\,dx = \ln\lvert x \rvert + C.

This is the missing case n=−1n = -1 of the power rule. It holds because

ddxln⁡∣x∣=1xfor every x≠0.\frac{d}{dx}\ln\lvert x \rvert = \frac{1}{x} \quad \text{for every } x \neq 0.

In plain words The integral of 1/x1/x is the natural logarithm, with an absolute value so that it also works for negative xx.

Example For x≠0x \neq 0 the integrand splits as

2x+5x=2+5x,\frac{2x + 5}{x} = 2 + \frac{5}{x},

so on either interval (0,∞)(0, \infty) or (−∞,0)(-\infty, 0)

∫2x+5x dx=2x+5ln⁡∣x∣+C.\int \frac{2x + 5}{x}\,dx = 2x + 5\ln\lvert x \rvert + C.

Exponential rules

Definition On R\mathbb{R},

∫ex dx=ex+C,∫ax dx=axln⁡a+C(a>0, a≠1).\int e^x\,dx = e^x + C, \qquad \int a^x\,dx = \frac{a^x}{\ln a} + C \quad (a > 0,\ a \neq 1).

More generally, for a constant k≠0k \neq 0,

∫ekx dx=1kekx+C.\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C.

In plain words The exponential exe^x is its own antiderivative. Other bases pick up a factor 1/ln⁡a1/\ln a.

Example On R\mathbb{R}

∫(ex+2x) dx=ex+2xln⁡2+C.\int (e^x + 2^x)\,dx = e^x + \frac{2^x}{\ln 2} + C.

Check this by differentiating, using the fact that the derivative of 2x2^x is 2xln⁡22^x\ln 2.

Trigonometric and inverse trigonometric rules

Definition With xx in radians, and each formula on any interval where its integrand is defined,

∫sin⁡x dx=−cos⁡x+C,∫cos⁡x dx=sin⁡x+C,∫sec⁡2x dx=tan⁡x+C,∫csc⁡2x dx=−cot⁡x+C,∫sec⁡xtan⁡x dx=sec⁡x+C,∫csc⁡xcot⁡x dx=−csc⁡x+C,\begin{aligned} \int \sin x\,dx &= -\cos x + C, & \int \cos x\,dx &= \sin x + C, \\ \int \sec^2 x\,dx &= \tan x + C, & \int \csc^2 x\,dx &= -\cot x + C, \\ \int \sec x\tan x\,dx &= \sec x + C, & \int \csc x\cot x\,dx &= -\csc x + C, \end{aligned}

and

∫dx1−x2=arcsin⁡x+C  on (−1,1),∫dx1+x2=arctan⁡x+C  on R.\int \frac{dx}{\sqrt{1 - x^2}} = \arcsin x + C \ \text{ on } (-1, 1), \qquad \int \frac{dx}{1 + x^2} = \arctan x + C \ \text{ on } \mathbb{R}.

In plain words Each formula is a derivative from the trigonometry lessons read backward. Watch the minus signs in the formulas for sine, cosecant and cotangent.

Example On (−1,1)(-1, 1),

∫(41+x2−31−x2)dx=4arctan⁡x−3arcsin⁡x+C.\int \left(\frac{4}{1 + x^2} - \frac{3}{\sqrt{1 - x^2}}\right)dx = 4\arctan x - 3\arcsin x + C.

Domains of the formulas

Definition Each rule gives the general antiderivative only on an interval contained in the domain of the integrand. If that domain splits into several disjoint intervals, the constant of integration may be chosen separately on each one.

In plain words Always know which interval you are working on, and do not let a formula jump across a point where the integrand is undefined.

Example The function sec⁡2x\sec^2 x is undefined where cos⁡x=0\cos x = 0. On each interval (−π/2+kπ,π/2+kπ)(-\pi/2 + k\pi, \pi/2 + k\pi) with kk an integer,

∫sec⁡2x dx=tan⁡x+Ck,\int \sec^2 x\,dx = \tan x + C_k,

and the constants CkC_k may differ. Likewise ln⁡x+C\ln x + C is an antiderivative of 1/x1/x only on (0,∞)(0, \infty), while ln⁡∣x∣+C\ln\lvert x \rvert + C also covers (−∞,0)(-\infty, 0).

Check yourself

  1. Find ∫(6x2−2x+7) dx\int (6x^2 - 2x + 7)\,dx.
Show the answer

The integral is 2x3−x2+7x+C2x^3 - x^2 + 7x + C on R\mathbb{R}. Differentiating gives back 6x2−2x+76x^2 - 2x + 7.

  1. Find ∫x2+1x dx\int \frac{x^2 + 1}{x}\,dx and say where your answer is valid.
Show the answer

For x≠0x \neq 0 the integrand is x+1/xx + 1/x, so

∫x2+1x dx=x22+ln⁡∣x∣+C,\int \frac{x^2 + 1}{x}\,dx = \frac{x^2}{2} + \ln\lvert x \rvert + C,

valid on (0,∞)(0, \infty) and on (−∞,0)(-\infty, 0), each with its own constant.

  1. Find ∫(3cos⁡x−2sec⁡2x) dx\int (3\cos x - 2\sec^2 x)\,dx.
Show the answer

The integral is 3sin⁡x−2tan⁡x+C3\sin x - 2\tan x + C, on any interval where cos⁡x≠0\cos x \neq 0.

How this lesson connects

It also uses elementary derivative formulas.