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Pure · Integration

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Integration by substitution

The formal version of the reverse chain rule — substitute u = g(x), replace dx using du = g'(x) dx, integrate in u, then convert back (or change the limits for a definite integral).

Pure · Integration

Integration by substitution

The formal version of the reverse chain rule — substitute u = g(x), replace dx using du = g'(x) dx, integrate in u, then convert back (or change the limits for a definite integral).

Why it works

Integration by recognition works when you can see the answer. Substitution is the systematic version for when you can't: you rename the awkward inner part as a new variable uu, and the whole integral simplifies into something standard.

The mechanics rest on one substitution for the differential. If u=g(x)u = g(x), then dudx=g′(x)⟹du=g′(x) dx.\frac{du}{dx} = g'(x) \quad\Longrightarrow\quad du = g'(x)\,dx. You treat dudx\dfrac{du}{dx} like a fraction here (legitimate inside an integral): wherever g′(x) dxg'(x)\,dx appears, replace it by dudu, and replace the inner g(x)g(x) by uu. Every xx must disappear — the new integral has to be entirely in uu — or the substitution isn't the right one.

Take ∫2x(x2+1)3 dx\int 2x(x^2 + 1)^3\,dx with u=x2+1u = x^2 + 1. Then du=2x dxdu = 2x\,dx, so 2x dx2x\,dx becomes dudu and (x2+1)3(x^2+1)^3 becomes u3u^3: ∫2x(x2+1)3 dx=∫u3 du=14u4+c=14(x2+1)4+c.\int 2x(x^2+1)^3\,dx = \int u^3\,du = \tfrac14 u^4 + c = \tfrac14(x^2 + 1)^4 + c. Often you'll need to rearrange the substitution to replace a stray xx too — e.g. from u=2x+1u = 2x + 1 you also get x=u−12x = \tfrac{u-1}{2} to deal with an extra factor of xx.

Definite integrals: change the limits. With a definite integral you have two clean choices — and the slick one is to convert the xx-limits into uu-limits using u=g(x)u = g(x), integrate in uu, and evaluate. Then you never convert back. For ∫022x(x2+1)3 dx\int_0^2 2x(x^2+1)^3\,dx with u=x2+1u = x^2 + 1: when x=0x = 0, u=1u = 1; when x=2x = 2, u=5u = 5; so it becomes ∫15u3 du=[14u4]15=625−14=156\int_1^5 u^3\,du = \big[\tfrac14 u^4\big]_1^5 = \tfrac{625 - 1}{4} = 156.