Guides / Integrals

u-substitution and integration by parts: the du you can't skip, and which part to call u

Most integrals past the basic table are done with one of two moves. Substitution undoes the chain rule, and integration by parts undoes the product rule. Both are short when the choice of u is right, and both have one piece of bookkeeping that gets dropped: the constant that comes with du, the limits of a definite integral, and which factor gets differentiated.

Short answer

Substitution: look for a function whose derivative is also in the integral, up to a constant. Call it u, write du = u′(x) dx, and replace every x, including the dx. If the derivative is off by a constant, divide by it: x dx = ½ du. In a definite integral, change the limits to values of u. Parts: ∫ u dv = uv − ∫ v du. Pick as u the factor that gets simpler when you differentiate it (a log, a power of x) and as dv the one you can integrate (an exponential, a sine or cosine). If the new integral is harder than the old one, swap them.

Why it works

The chain rule says ddx F(u(x)) = F′(u(x)) · u′(x). Read backwards, any integral of the form ∫ f(u(x)) u′(x) dx is F(u(x)) + C, where F is an antiderivative of f. Writing du = u′(x) dx is just a way of spotting that shape: the u′(x) dx has to be there, and if it isn't all there, the substitution doesn't work.

The product rule says (uv)′ = u′v + uv′. Integrate both sides and rearrange, and you get ∫ uv′ dx = uv − ∫ u′v dx, which is the parts formula. It doesn't finish an integral by itself. It trades one integral for another, and the choice of u is what makes the trade a good one.

Example 1: substitution, with a constant to fix

Find ∫ x cos(x²) dx.

u = x², du = 2x dx, so x dx = 12 duThe inside of the cosine has derivative 2x, and x dx is already in the integral. Only the 2 is missing.
∫ cos u · 12 du = 12 sin u + CEvery x is gone, including the dx.
= 12 sin(x²) + CCheck: the derivative is ½ · cos(x²) · 2x = x cos(x²).

Example 2: a definite integral, limits in u

Evaluate ∫₀¹ x(x² + 1)³ dx.

u = x² + 1, x dx = 12 du; x = 0 → u = 1, x = 1 → u = 2Change the limits at the same time as the variable.
12∫₁² u³ du = 12 [u⁴4]₁²The integral is now entirely in u, limits included.
= 12 · 16 − 14 = 158No need to go back to x: the limits are already in u.

Example 3: integration by parts

Find ∫ x e^2x dx.

u = x, dv = e^2x dx; du = dx, v = 12e^2xDifferentiating x makes it 1, and e^(2x) is easy to integrate.
∫ x e^2x dx = 12x e^2x − ∫ 12e^2x dxuv − ∫ v du. The new integral has no x in front.
= 12x e^2x − 14e^2x + CCheck: the derivative is ½e^(2x) + x e^(2x) − ½e^(2x) = x e^(2x).

Example 4: parts twice

Find ∫ x² e^x dx.

u = x², dv = e^x dx: ∫ x² e^x dx = x² e^x − ∫ 2x e^x dxOne round of parts takes the power of x down by one.
∫ 2x e^x dx = 2x e^x − ∫ 2e^x dx = 2x e^x − 2e^xA second round, again with the power of x as u.
= x² e^x − 2x e^x + 2e^x + C = e^x(x² − 2x + 2) + CKeep the second integral in parentheses when you subtract it, or the sign of the 2e^x comes out wrong.

Common mistakes, and the exact line they happen on

1. Dropping the constant that comes with du

The integral from Example 1.

u = x², du = 2x dxRight so far.
∫ x cos(x²) dx = ∫ cos u du = sin(x²) + CThis is the mistake. x dx was replaced by du, but du is 2x dx, twice as much. Differentiating catches it: the derivative of sin(x²) is 2x cos(x²), twice the integrand. Divide by the 2: x dx = ½ du, and the answer is ½ sin(x²) + C.

2. Keeping the x-limits after substituting

The definite integral from Example 2.

12∫ u³ du, with u = x² + 1Right so far.
12 [u⁴4]₀¹ = 18This is the mistake. 0 and 1 are values of x. When x runs from 0 to 1, u = x² + 1 runs from 1 to 2, and those limits give 15/8. Either change the limits, or put x back before you evaluate.

3. Choosing u and dv the wrong way round

The integral from Example 3.

∫ x e^2x dxRight so far.
u = e^2x, dv = x dx: = x²2e^2x − ∫ x² e^2x dxThis is the mistake. The equation is true, but the new integral has x² where the old one had x, so it is harder, and each further round makes it worse. Make u the factor that gets simpler when differentiated, here x, and the new integral has no x at all.

Practice

Find ∫ 3x²(x³ + 1)⁴ dx.
Show answer
u = x³ + 1, du = 3x² dx exactly, so the answer is (x³ + 1)⁵5 + C.
Find ∫ e^x1 + e^x dx.
Show answer
u = 1 + e^x, du = e^x dx, so ∫ duu = ln(1 + e^x) + C. No absolute value is needed, because 1 + e^x > 0.
Find ∫ x sin x dx.
Show answer
Parts with u = x, dv = sin x dx, v = −cos x: −x cos x + ∫ cos x dx = −x cos x + sin x + C.
Evaluate ∫₁^e ln x dx.
Show answer
Parts with u = ln x, dv = dx: [x ln x − x]₁^e = (e − e) − (0 − 1) = 1.
Find ∫ arctan x dx.
Show answer
Parts with u = arctan x, dv = dx, then the substitution w = 1 + x² in what is left: x arctan x − ½ ln(1 + x²) + C.