Integration by Parts

The product rule, rearranged

The product rule says that for two functions u and v of x, (uv)' = u'v + uv'. Take u'v to the other side: uv' = (uv)' − u'v.

Now integrate both sides. Integrating (uv)' gives back uv, so $\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx$∫ uv' dx = uv − ∫ u'v dx. Writing dv for v' dx and du for u' dx, this is $\int udv=uv-\int vdu$∫ u dv = uv − ∫ v du. This is integration by parts.

It does not finish an integral by itself. It trades $\int udv$∫ u dv for a different integral, $\int vdu$∫ v du, and the trade is worth making only when the new integral is easier than the old one.

The curve $v=u^2$v = u² drawn against u, inside the box from the origin to the corner (1.8, 3.24). The region under the curve is $\int vdu=1.94$∫ v du = 1.94 and the region to its left is $\int udv=3.89$∫ u dv = 3.89; together they fill the box, uv = 1.8 × 3.24 = 5.83. Drag the corner: the two regions always fill the box, which is $\int udv=uv-\int vdu$∫ u dv = uv − ∫ v du.

A first example

Take $\int x{e}^{x}dx$∫ x eˣ dx. Let u = x and $dv=e^{x}dx$dv = eˣ dx. Then du = dx, and $v=e^x$v = eˣ, because $e^x$eˣ differentiates to itself.

The formula gives $\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+c$∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + c. The x has gone from the new integral, which is why it was easy.

Check by differentiating, with the product rule on $x{e}^x$x eˣ: $e^x+x{e}^x-e^x=x{e}^x$eˣ + x eˣ − eˣ = x eˣ, the integrand.

Choosing u

Choose u as the part that becomes simpler when it is differentiated, and dv as a part you can integrate. A power of x is the usual u: differentiating it lowers the power.

The other choice for $\int x{e}^{x}dx$∫ x eˣ dx is $u=e^x$u = eˣ and dv = x dx. Then $du=e^{x}dx$du = eˣ dx and $v=\frac{x^2}{2}$v = x²/2, and the formula gives $\frac{x^2{e}^x}{2}-\int \left(\frac{x^2}{2}\right)e^{x}dx$x²eˣ/2 − ∫ (x²/2) eˣ dx. The new integral has $x^2$x² where the old one had x, so it is harder, and the trade has gone the wrong way.

In $\int x\cos xdx$∫ x cos x dx, take u = x and dv = cos x dx, so du = dx and v = sin x. Then $\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+c$∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + c, because −cos x is an antiderivative of sin x. Check: sin x + x cos x − sin x = x cos x.

For $\int x{e}^{x}dx$∫ x eˣ dx this starts with $u=e^x$u = eˣ and dv = x dx, which leaves $\int \left(\frac{x^2}{2}\right)e^{x}dx$∫ (x²/2) eˣ dx, harder than the integral started with. Choose u = x instead, and the integral left is $\int e^{x}dx$∫ eˣ dx. The other two integrals at the top work the same way.

When dv is just dx

The integrand ln x is not a product, and no rule so far integrates it. Write it as ln x × 1 and take u = ln x and dv = dx. Then $du=\left(\frac{1}{x}\right)dx$du = (1/x) dx and v = x.

The formula gives $\int \ln xdx=x\ln x-\int x\times \left(\frac{1}{x}\right)dx=x\ln x-\int 1dx=x\ln x-x+c$∫ ln x dx = x ln x − ∫ x × (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + c. Check: the derivative of x ln x is $\ln x+x\times \left(\frac{1}{x}\right)=\ln x+1$ln x + x × (1/x) = ln x + 1, and taking away the derivative of x leaves ln x.

Two rounds

For $\int x^2{e}^{x}dx$∫ x²eˣ dx, take $u=x^2$u = x² and $dv=e^{x}dx$dv = eˣ dx. Then du = 2x dx and $v=e^x$v = eˣ, so $\int x^2{e}^{x}dx=x^2{e}^x-\int 2x{e}^{x}dx$∫ x²eˣ dx = x²eˣ − ∫ 2x eˣ dx. The power has dropped from 2 to 1, but one integral by parts remains.

The first example gives $\int 2x{e}^{x}dx=2\left(x{e}^x-e^x\right)$∫ 2x eˣ dx = 2(x eˣ − eˣ). So $\int x^2{e}^{x}dx=x^2{e}^x-2x{e}^x+2e^x+c=\left(x^2-2x+2\right)e^x+c$∫ x²eˣ dx = x²eˣ − 2x eˣ + 2eˣ + c = (x² − 2x + 2)eˣ + c. Check: the derivative is $\left(2x-2\right)e^x+\left(x^2-2x+2\right)e^x=x^2{e}^x$(2x − 2)eˣ + (x² − 2x + 2)eˣ = x²eˣ.

Definite integrals

Evaluate the antiderivative at both ends as usual. From 0 to 1, $x{e}^x-e^x$x eˣ − eˣ gives (e − e) − (0 − 1) = 1, so the integral of $x{e}^x$x eˣ from 0 to 1 is 1. 1000 midpoint strips add to 0.9999998.

From 0 to $\pi$π, x sin x + cos x gives $\left(\pi \times 0-1\right)-\left(0+1\right)=-2$(π × 0 − 1) − (0 + 1) = −2, so the integral of x cos x from 0 to $\pi$π is −2: more of the region lies below the axis, where cos x is negative, than above it. 1000 midpoint strips add to −1.999999.

The usual mistakes

Adding the second integral instead of subtracting it. $x{e}^x+e^x$x eˣ + eˣ differentiates to $2e^x+x{e}^x$2eˣ + x eˣ, not $x{e}^x$x eˣ.

Integrating each factor separately. $\frac{x^2{e}^x}{2}$x²eˣ/2 differentiates to $x{e}^x+\frac{x^2{e}^x}{2}$x eˣ + x²eˣ/2, because integrals do not split over a product; parts exists to handle the second term.

Choosing $u=e^x$u = eˣ for $\int x{e}^{x}dx$∫ x eˣ dx. $e^x$eˣ never gets simpler when it is differentiated, and the leftover integral gains a power of x.

Losing a sign in $\int \sin xdx$∫ sin x dx. It is −cos x, so $x\sin x-\int \sin xdx$x sin x − ∫ sin x dx is x sin x + cos x; x sin x − cos x differentiates to 2 sin x + x cos x.

A timber kiln

In the application below, moisture leaves timber at a rate that is a power of time times a falling exponential. The power is differentiated and the exponential integrated, so the integral left over is an exponential alone.