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Integration Rules — Definition, Formula & Examples

Integration rules are a set of standard formulas and properties that allow you to compute integrals of common functions without deriving each result from scratch. They include the power rule, constant multiple rule, sum/difference rule, and formulas for exponential, logarithmic, and trigonometric integrands.

Integration rules are identities governing the integral operator, derived from corresponding differentiation rules applied in reverse. For a continuous function ff on an interval II, these rules express f(x)dx\int f(x)\,dx in closed form or reduce it to simpler integrals. Key rules include: (1) xndx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C for n1n \neq -1; (2) kf(x)dx=kf(x)dx\int k\,f(x)\,dx = k\int f(x)\,dx; (3) [f(x)±g(x)]dx=f(x)dx±g(x)dx\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx; (4) exdx=ex+C\int e^x\,dx = e^x + C; (5) 1xdx=lnx+C\int \frac{1}{x}\,dx = \ln|x| + C.

Key Formula

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$$ $$\int k\,f(x)\,dx = k\int f(x)\,dx$$ $$\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$$ $$\int e^x\,dx = e^x + C \qquad \int \frac{1}{x}\,dx = \ln|x| + C$$ $$\int \sin x\,dx = -\cos x + C \qquad \int \cos x\,dx = \sin x + C
Where:
  • nn = Any real exponent except −1
  • kk = A constant factor
  • CC = The arbitrary constant of integration
  • f(x),g(x)f(x), g(x) = Continuous functions of x

How It Works

Each integration rule tells you the antiderivative of a particular type of expression. To integrate a complicated function, you first break it apart using the sum/difference rule and the constant multiple rule, then apply specific formulas (power rule, exponential rule, trig rules) to each piece. For instance, facing (3x45sinx)dx\int (3x^4 - 5\sin x)\,dx, you split it into 3x4dx5sinxdx3\int x^4\,dx - 5\int \sin x\,dx and apply the power rule and the sine rule separately. When these basic rules are not enough, you move to advanced techniques like substitution or integration by parts, which themselves rely on these foundational identities.

Worked Example

Problem: Evaluate (6x2+4ex3x)dx\int (6x^2 + 4e^x - \frac{3}{x})\,dx.
Step 1: Split using sum/difference and constant multiple rules: Distribute the integral across each term and pull out constants.
6x2dx+4exdx31xdx6\int x^2\,dx + 4\int e^x\,dx - 3\int \frac{1}{x}\,dx
Step 2: Apply the power rule to $x^2$: With n=2n = 2, the power rule gives x33\frac{x^3}{3}.
6x33=2x36 \cdot \frac{x^3}{3} = 2x^3
Step 3: Apply the exponential rule: The integral of exe^x is exe^x.
4ex=4ex4 \cdot e^x = 4e^x
Step 4: Apply the reciprocal rule: The integral of 1x\frac{1}{x} is lnx\ln|x|.
3lnx=3lnx-3 \cdot \ln|x| = -3\ln|x|
Step 5: Combine and add the constant: Write the full antiderivative with a single constant CC.
2x3+4ex3lnx+C2x^3 + 4e^x - 3\ln|x| + C
Answer: 2x3+4ex3lnx+C2x^3 + 4e^x - 3\ln|x| + C

Another Example

Problem: Evaluate (5cosx+x1/2)dx\int (5\cos x + x^{-1/2})\,dx.
Step 1: Separate the integral: Split into two integrals using the sum rule.
5cosxdx+x1/2dx5\int \cos x\,dx + \int x^{-1/2}\,dx
Step 2: Integrate $\cos x$: The cosine rule gives sinx\sin x.
5sinx5\sin x
Step 3: Apply the power rule with $n = -\frac{1}{2}$: Add 1 to the exponent to get 12\frac{1}{2}, then divide by 12\frac{1}{2}.
x1/21/2=2x\frac{x^{1/2}}{1/2} = 2\sqrt{x}
Step 4: Combine: Write the complete antiderivative.
5sinx+2x+C5\sin x + 2\sqrt{x} + C
Answer: 5sinx+2x+C5\sin x + 2\sqrt{x} + C

Why It Matters

Integration rules are the backbone of every Calculus I and Calculus II course; nearly every homework problem and exam question begins with them. Engineers use these formulas to compute areas, volumes, work, and fluid pressure. Physicists rely on them to derive equations of motion, electric potential, and probability distributions from continuous models.

Common Mistakes

Mistake: Forgetting to add the constant of integration CC in indefinite integrals.
Correction: Every indefinite integral represents a family of functions differing by a constant. Always write +C+ C unless you are evaluating a definite integral with limits.
Mistake: Applying the power rule when n=1n = -1, producing x00\frac{x^0}{0}.
Correction: The power rule excludes n=1n = -1. For x1dx\int x^{-1}\,dx, use lnx+C\ln|x| + C instead.

Related Terms