Integration by Substitution

The chain rule leaves a factor

Differentiate ${\left(x^2+1\right)}^6$(x² + 1)⁶ by the chain rule. The outside power gives $6{\left(x^2+1\right)}^5$6(x² + 1)⁵, and the inside $x^2+1$x² + 1 has derivative 2x, which multiplies the result: the derivative is $6{\left(x^2+1\right)}^5\times 2x$6(x² + 1)⁵ × 2x.

So whenever an integral is a function of some inside expression, multiplied by the derivative of that inside, it is a chain-rule result waiting to be run backwards. Substitution is the method for running it backwards.

Spot the factor, name the inside

Take $\int 2x{\left(x^2+1\right)}^{5}dx$∫ 2x(x² + 1)⁵ dx. The bracket $x^2+1$x² + 1 is an inside expression, and its derivative 2x stands beside it as a factor. That pairing is the signal.

Name the inside u: let $u=x^2+1$u = x² + 1. The bracket ${\left(x^2+1\right)}^5$(x² + 1)⁵ becomes $u^5$u⁵, an ordinary power.

Trade 2x dx for du

Differentiating $u=x^2+1$u = x² + 1 gives $\frac{du}{dx}=2x$du/dx = 2x, which as differentials is du = 2x dx. The integral holds exactly 2x dx, so it can be replaced by du, and every x is gone: $\int 2x{\left(x^2+1\right)}^{5}dx=\int u^{5}du$∫ 2x(x² + 1)⁵ dx = ∫ u⁵ du.

Integrate the power: $\int u^{5}du=\frac{u^6}{6}+c$∫ u⁵ du = u⁶/6 + c. Then write $x^2+1$x² + 1 back in place of u, because u was only a name: $\int 2x{\left(x^2+1\right)}^{5}dx=\frac{{\left(x^2+1\right)}^6}{6}+c$∫ 2x(x² + 1)⁵ dx = (x² + 1)⁶/6 + c.

Check by differentiating. The chain rule gives $6{\left(x^2+1\right)}^5\times 2x/6=2x{\left(x^2+1\right)}^5$6(x² + 1)⁵ × 2x / 6 = 2x(x² + 1)⁵, the integrand.

x along the top line and $u=x^2$u = x² along the bottom. The interval from x = 1.2 to 1.4, 0.2 wide, lands on u from 1.44 to 1.96, which is 0.52 wide: 2.6 times as wide, close to $\frac{du}{dx}=2x=2.4$du/dx = 2x = 2.4. That stretch is the factor du = 2x dx uses up. Slide the interval to x = 0.4, where the two widths match.

Three more

For $\int 3x^2{\left(x^3+2\right)}^{4}dx$∫ 3x²(x³ + 2)⁴ dx, let $u=x^3+2$u = x³ + 2. Then $du=3x^{2}dx$du = 3x² dx, which is exactly the factor present, so the integral is $\int u^{4}du=\frac{u^5}{5}+c=\frac{{\left(x^3+2\right)}^5}{5}+c$∫ u⁴ du = u⁵/5 + c = (x³ + 2)⁵/5 + c. Check: $5{\left(x^3+2\right)}^4\times 3x^2/5=3x^2{\left(x^3+2\right)}^4$5(x³ + 2)⁴ × 3x² / 5 = 3x²(x³ + 2)⁴.

A factor that is out by a constant still works. In $\int x{\left(x^2+1\right)}^{5}dx$∫ x(x² + 1)⁵ dx, du = 2x dx but only x dx is present. Since x dx = ½ du, the integral is $\frac{1}{2}\int u^{5}du=\frac{u^6}{12}+c=\frac{{\left(x^2+1\right)}^6}{12}+c$½ ∫ u⁵ du = u⁶/12 + c = (x² + 1)⁶/12 + c. Check: $6{\left(x^2+1\right)}^5\times 2x/12=x{\left(x^2+1\right)}^5$6(x² + 1)⁵ × 2x / 12 = x(x² + 1)⁵.

A root is a power too. In $\int 2x\sqrt{x^2+9}dx$∫ 2x√(x² + 9) dx, let $u=x^2+9$u = x² + 9, so du = 2x dx and the integral is $\int \sqrt{u}du$∫ √u du, the integral of u to the power ½. Raising the power to $\frac{3}{2}$3/2 and dividing by $\frac{3}{2}$3/2 gives $\frac{2}{3}u\sqrt{u}+c=\frac{2}{3}\left(x^2+9\right)\sqrt{x^2+9}+c$⅔ u√u + c = ⅔ (x² + 9)√(x² + 9) + c. Check: the derivative of the power $\frac{3}{2}$3/2 brings down $\frac{3}{2}$3/2, which cancels the ⅔, and leaves $\sqrt{x^2+9}$√(x² + 9) times the inside derivative 2x.

When substitution fails

In $\int {\left(x^2+1\right)}^{5}dx$∫ (x² + 1)⁵ dx there is no factor 2x. Letting $u=x^2+1$u = x² + 1 still gives du = 2x dx, so $dx=\frac{du}{2x}$dx = du/(2x), and that leaves an x inside an integral in u. An x left over means the substitution does not work; here the bracket has to be multiplied out instead.

Choosing the wrong part as u fails the same way. In $\int 2x{\left(x^2+1\right)}^{5}dx$∫ 2x(x² + 1)⁵ dx, u = 2x gives du = 2 dx, and the bracket ${\left(x^2+1\right)}^5$(x² + 1)⁵ is still written in x.

A definite integral, two ways

Find the integral of $3x^2{\left(x^3+2\right)}^4$3x²(x³ + 2)⁴ from x = 0 to x = 1. The lesson’s way is to finish in x, then evaluate: $\left[\frac{{\left(x^3+2\right)}^5}{5}\right]$[(x³ + 2)⁵/5] from 0 to 1 is $\frac{3^5}{5}-\frac{2^5}{5}=\frac{243-32}{5}=\frac{211}{5}=42.2$3⁵/5 − 2⁵/5 = (243 − 32)/5 = 211/5 = 42.2.

The other way changes the limits into values of u and never returns to x. At x = 0, $u=0^3+2=2$u = 0³ + 2 = 2; at x = 1, $u=1^3+2=3$u = 1³ + 2 = 3. So the integral is the integral of $u^4$u⁴ from u = 2 to u = 3, which is $\frac{3^5}{5}-\frac{2^5}{5}=\frac{211}{5}$3⁵/5 − 2⁵/5 = 211/5 again.

Check it by strips: 1000 midpoint rectangles from 0 to 1 add to 42.19994.

The usual mistakes

Taking the inside’s derivative as u. In $\int 2x{\left(x^2+1\right)}^{5}dx$∫ 2x(x² + 1)⁵ dx the 2x becomes part of du; u is the bracket $x^2+1$x² + 1.

Leaving an x beside the bracket. $\frac{x{\left(x^2+1\right)}^6}{6}+c$x(x² + 1)⁶/6 + c keeps the 2x that du = 2x dx has already used up.

Skipping the divide. ${\left(x^2+1\right)}^6+c$(x² + 1)⁶ + c differentiates to $12x{\left(x^2+1\right)}^5$12x(x² + 1)⁵, six times the integrand.

Keeping the x limits once the integral is in u. The integral of $u^4$u⁴ from 0 to 1 is $\frac{1}{5}$1/5; the limits in u are 2 and 3.

A sluice gate

In the application below, the flow through a sluice has $n^2+144$n² + 144 under a root. Substituting for it turns the root into a power, and the limits are carried into the new variable, where they come out as whole-number square roots.