Integrating Powers

The rule, and why it works

To integrate $x^n$xⁿ, add one to the power, then divide by the new power: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, for every n except −1.

Differentiating the answer proves it. The power n + 1 comes down in front and the power drops back to n, giving $\frac{\left(n+1\right)x^n}{n+1}$(n + 1)xⁿ/(n + 1). The n + 1 on top cancels the n + 1 underneath, leaving $x^n$xⁿ. Nothing in that argument needs n to be a positive whole number, so the rule works for negative powers and fractional powers too. The one thing it needs is that n + 1 is not 0.

Negative powers

Take $x^{-3}$x⁻³. Adding one to −3 gives −2, so the power goes up to −2, not down to −4. Divide by the new power, −2: $\int x^{-3}dx=\frac{x^{-2}}{-2}+C=-\frac{1}{2x^2}+C$∫ x⁻³ dx = x⁻²/(−2) + C = −1/(2x²) + C. Check: $-\frac{1}{2}{x}^{-2}$−½x⁻² differentiates to $\left(-\frac{1}{2}\right)\left(-2\right)x^{-3}=x^{-3}$(−½)(−2)x⁻³ = x⁻³.

A fraction with a power of x underneath is a negative power, so write it that way first. $\frac{1}{x^2}=x^{-2}$1/x² = x⁻², which integrates to $\frac{x^{-1}}{-1}=-\frac{1}{x}+C$x⁻¹/(−1) = −1/x + C. Check: $-x^{-1}$−x⁻¹ differentiates to $x^{-2}$x⁻², which is $\frac{1}{x^2}$1/x².

The gold curve is $y=-\frac{1}{x}$y = −1/x and the plain curve is $y=\frac{1}{x^2}$y = 1/x², for x > 0. The tangent to $y=-\frac{1}{x}$y = −1/x at (1, −1) is y = x − 2, with gradient 1, and $\frac{1}{x^2}$1/x² is 1 at x = 1. At x = 2, $-\frac{1}{x}$−1/x has gradient $\frac{1}{4}$1/4, and $\frac{1}{x^2}$1/x² is 0.25 there.

Fractional powers

A square root is the power ½: $\sqrt{x}=x^{\frac{1}{2}}$√x = x^(1/2). Adding one gives the power $\frac{3}{2}$3/2, and dividing by $\frac{3}{2}$3/2 is the same as multiplying by $\frac{2}{3}$2/3. So $\int \sqrt{x}dx=\frac{2x^{\frac{3}{2}}}{3}+C$∫ √x dx = 2x^(3/2)/3 + C. Check: differentiating brings down $\frac{3}{2}$3/2, and $\left(\frac{2}{3}\right)\left(\frac{3}{2}\right)=1$(2/3)(3/2) = 1, leaving $x^{\frac{1}{2}}$x^(1/2), which is $\sqrt{x}$√x.

One over a root is a negative fractional power: $\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$1/√x = x^(−1/2). Adding one gives the power ½, and dividing by ½ doubles: $\int \frac{1}{\sqrt{x}}dx=2\sqrt{x}+C$∫ 1/√x dx = 2√x + C. Check: $2\sqrt{x}$2√x differentiates to $2\times \frac{1}{2}{x}^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}$2 × ½ x^(−1/2) = 1/√x.

A constant is a power too. $5=5x^0$5 = 5x⁰, which integrates to $\frac{5x^1}{1}=5x$5x¹/1 = 5x.

Rewrite as powers first

The rule handles one power of x at a time, so brackets and quotients are turned into sums of powers before integrating. ${\left(x+2\right)}^2=x^2+4x+4$(x + 2)² = x² + 4x + 4, which integrates to $\frac{x^3}{3}+2x^2+4x+C$x³/3 + 2x² + 4x + C.

A quotient with a single term underneath splits into separate terms. $\frac{x^3+1}{x^2}=\frac{x^3}{x^2}+\frac{1}{x^2}=x+x^{-2}$(x³ + 1)/x² = x³/x² + 1/x² = x + x⁻², which integrates to $\frac{x^2}{2}-\frac{1}{x}+C$x²/2 − 1/x + C. Differentiating that gives $x+x^{-2}$x + x⁻² back.

Signs need care when the power is negative. $4x^3-\frac{6}{x^3}=4x^3-6x^{-3}$4x³ − 6/x³ = 4x³ − 6x⁻³. The first term gives $x^4$x⁴. The second gives $-6\times \frac{x^{-2}}{-2}=3x^{-2}$−6 × x⁻²/(−2) = 3x⁻², so the integral is $x^4+\frac{3}{x^2}+C$x⁴ + 3/x² + C. Check: $3x^{-2}$3x⁻² differentiates to $-6x^{-3}$−6x⁻³, the term that was there.

Why n = −1 is left out

Put n = −1 into the rule and it asks for $\frac{x^0}{0}$x⁰/0, a division by zero. The formula breaks, and no repair of it can work, because no power of x differentiates to $x^{-1}$x⁻¹.

Here is why. A power $x^k$xᵏ differentiates to $k{x}^{k-1}$k xᵏ⁻¹. To land on the power −1 the old power would have to be k = 0, and then the multiplier in front is that same 0: $x^0=1$x⁰ = 1 differentiates to 0, not to $x^{-1}$x⁻¹.

Yet $\frac{1}{x}$1/x does have an antiderivative. The region under $y=\frac{1}{x}$y = 1/x from x = 1 to x = 6 has a definite size, and measured from 1 up to a moving right edge at x, that area grows at the rate $\frac{1}{x}$1/x, the height of the curve there. So the antiderivative exists; it is simply not a power of x. It is a new function, met in its own lesson. Powers close to −1 cause no trouble: $x^{-0.9}$x^(−0.9) integrates to $\frac{x^{0.1}}{0.1}=10x^{0.1}+C$x^(0.1)/0.1 = 10x^(0.1) + C.

The curve $y=\frac{1}{x}$y = 1/x, with the region under it from x = 1 to x = 6 shaded. The region has an area, so $\frac{1}{x}$1/x has an antiderivative, but no power of x is that antiderivative.

The usual mistakes

Moving a negative power the wrong way. Adding one to −3 gives −2. Writing $\frac{x^{-4}}{-4}$x⁻⁴/(−4) for the integral of $x^{-3}$x⁻³ subtracts one instead.

Dividing by a fraction the wrong way up. Dividing by $\frac{3}{2}$3/2 means multiplying by $\frac{2}{3}$2/3, so $\sqrt{x}$√x integrates to $\frac{2x^{\frac{3}{2}}}{3}$2x^(3/2)/3, not $\frac{3x^{\frac{3}{2}}}{2}$3x^(3/2)/2.

Integrating the top and the bottom of a quotient separately. $\frac{x^3+1}{x^2}$(x³ + 1)/x² must be split into $x+x^{-2}$x + x⁻² first; integrating the numerator and the denominator on their own and dividing gives a function that does not differentiate back.

Using the rule on $\frac{1}{x}$1/x. It gives $\frac{x^0}{0}$x⁰/0, which has no meaning. $\frac{1}{x}$1/x needs the new function from its own lesson.