The Power Rule for Integration — Moosa Academy

Add one to the exponent, divide by the new exponent, add $C$ . That single instruction covers more cases than any other rule in integration — plain powers, coefficients, roots, fractions and negative exponents all fall to it. There is exactly one exponent it cannot touch.

Theorem The power rule

$\displaystyle \int x^{n}\,dx = \frac{x^{n+1}}{n+1} + C$ , for any constant $n \neq -1$ .

Raise the exponent by one.

Divide by the new exponent.

Add the constant

C

Three steps, in that order, every time. The whole of this lesson is one rule applied to different-looking inputs.

Note The two conditions

The variable must sit in the base. The rule handles $x^{n}$ , where $x$ is the base and $n$ is a constant. Something like $2^{x}$ has the variable in the exponent instead, and needs a different rule entirely.

The exponent must not be $-1$ . Adding one to $-1$ gives $0$ , and the rule would then divide by $0$ . That single case is why $\ln x$ exists.

Case 1 A plain power

\int x^{2}\,dx

: the exponent

2

becomes

3

, and

3

goes underneath

$\int x^{2}\,dx = \dfrac{x^{3}}{3} + C$

\int x^{5}\,dx

: the exponent

5

becomes

6

$\int x^{5}\,dx = \dfrac{x^{6}}{6} + C$

The new exponent appears twice — once upstairs and once downstairs. Forgetting the denominator is the most common slip.

Case 2 A constant in front

Any constant multiplier comes straight out of the integral, untouched. Integrate the power, then tidy up the numbers.

\int 6x^{2}\,dx = 6 \int x^{2}\,dx = 6 \cdot \dfrac{x^{3}}{3}

$= 2x^{3} + C$

\int 4x^{3}\,dx = 4 \cdot \dfrac{x^{4}}{4}

, and the fours cancel

$= x^{4} + C$

\int \dfrac{2}{3}x^{2}\,dx = \dfrac{2}{3} \cdot \dfrac{x^{3}}{3}

$= \dfrac{2x^{3}}{9} + C$

A fraction out front is still just a constant. Pull it out, integrate, then multiply the two denominators together.

Case 3 Roots and fractional exponents

A root is not a special case. Rewrite it as a power and the same three steps apply.

\sqrt{x} = x^{1/2}

, and

\tfrac{1}{2} + 1 = \tfrac{3}{2}

\int \sqrt{x}\,dx = \dfrac{x^{3/2}}{3/2}

, and dividing by

\tfrac{3}{2}

means multiplying by

\tfrac{2}{3}

$= \dfrac{2}{3}x^{3/2} + C$

\sqrt[3]{x} = x^{1/3}

, and

\tfrac{1}{3} + 1 = \tfrac{4}{3}

$\int \sqrt[3]{x}\,dx = \dfrac{3}{4}x^{4/3} + C$

The only extra work is fraction arithmetic. The rule itself does not change.

Case 4 Negative exponents

\int x^{-2}\,dx

: the exponent

-2

becomes

-1

, which also goes underneath

= \dfrac{x^{-1}}{-1} = -x^{-1}

$= -\dfrac{1}{x} + C$

\int x^{-3}\,dx = \dfrac{x^{-2}}{-2}

$= -\dfrac{1}{2x^{2}} + C$

Watch the signs. A negative new exponent lands in the denominator as a negative number, and that minus sign belongs to the whole term.

Note The constant function, again

A bare constant has no $x$ beside it, but you can put one there: $x^{0} = 1$ , so $5 = 5x^{0}$ .

\int 5\,dx = \int 5x^{0}\,dx

The exponent

0

becomes

1

, and

1

goes underneath

= 5 \cdot \dfrac{x^{1}}{1}

$= 5x + C$

The same answer the rectangles gave. The rule from the previous lesson was never a separate rule — it is the power rule with $n = 0$ .

Note The one exponent it cannot handle

Try

n = -1

: the new exponent is

-1 + 1 = 0

The rule would give

\dfrac{x^{0}}{0}

Division by zero, so the rule does not apply

This gap is not a flaw in the rule. $\int \dfrac{1}{x}\,dx$ has an answer, but it is a different kind of function — the natural logarithm — and it gets a lesson of its own.

Summary

$\int x^{n}\,dx = \dfrac{x^{n+1}}{n+1} + C$ : raise the exponent, divide by the new one, add $C$ .
The variable must be the base, not the exponent.
Any constant multiplier comes outside the integral untouched.
Roots are fractional exponents, so they need no special treatment.
Negative exponents work too — the signs just need care.
$n = -1$ would divide by zero, and that case belongs to $\ln x$ .