Choosing u for U-Substitution: How to Pick u in an Integral

To pick u in u-substitution, set u equal to the inside function of a composite whose derivative also appears as a factor in the integrand, up to a constant you can move outside the integral. In practice, look for the expression under a root or power, the denominator of a fraction, the exponent on e, or the angle inside a trig function, then set u to that piece and check that its derivative is present. If the leftover part of the integrand matches the derivative of your chosen u, you picked correctly; if it does not, try a different inside piece before you commit.

Here is the one move that unsticks you on a blank page: set u equal to the inside function whose derivative also shows up in the integrand (up to a constant you can factor out). If picking that inside piece makes the leftover part look like its derivative, you chose right.

Follow this in order every time:

1. Scan for a composite f(g(x)) — a function tucked inside another function. Set u = the inside piece g(x).

2. Compute du = g'(x) dx and check that g'(x) is actually there as a factor. A constant multiple is fine — you can slide constants across the integral sign.

3. If nothing on the outside looks like the derivative of your inside piece, pick a different u before you commit.

Four pattern cues tell you what the inside piece usually is:

• u = the expression under a root or raised to a power.

• u = the denominator of a fraction.

• u = the exponent sitting on e.

• u = the angle inside a trig function like sin, cos, or tan.

Once u is chosen, rewrite dx in terms of du, integrate in u, then back-substitute so your answer is in x again:

\int f(u)\,du = F(u) + C = F(g(x)) + C

How do you pick u in u-substitution?

\int f(g(x))\,g'(x)\,dx \;\xrightarrow{\;u=g(x)\;}\; \int f(u)\,du

Worked example: choosing u for the integral of 2x times (x squared plus 1) to the fifth

Evaluate the integral of 2x·(x² + 1)⁵ with respect to x.

Step 1 — Spot the composite. The (x² + 1)⁵ is a fifth power of an inside piece, so the inside is x² + 1. That is a power-of cue, so set u there:

\displaystyle u = x^2 + 1

Step 2 — Differentiate and check the factor. The derivative of the inside is 2x, and 2x is sitting right there in the integrand, so this substitution will work:

du = 2x\,dx

Step 3 — Rewrite the whole integral in u. The 2x dx becomes du, and (x² + 1)⁵ becomes u⁵:

\int 2x\,(x^2+1)^5\,dx = \int u^5\,du

Step 4 — Integrate in u using the power rule:

\int u^5\,du = \frac{u^6}{6} + C

Step 5 — Back-substitute u = x² + 1 so the answer is in x:

\frac{(x^2+1)^6}{6} + C

Verify by differentiating: the chain rule gives one-sixth times six times (x² + 1) to the fifth times 2x, which is exactly 2x(x² + 1)⁵ — the original integrand, so the answer checks out.

\frac{d}{dx}\left[\frac{(x^2+1)^6}{6}\right] = 2x\,(x^2+1)^5

Another worked example

Evaluate the integral of x times cosine of (x squared). Step 1: the angle inside the cosine is x squared, which is the trig-angle cue, so set u equal to x squared. Step 2: the derivative of x squared is 2x, and the integrand has an x, which is 2x up to the constant one-half, so the substitution works. Step 3: since du equals 2x dx, the x dx in the integrand equals one-half du, so the integral becomes one-half times the integral of cosine u du. Step 4: integrate to get one-half times sine u, then back-substitute to get one-half times sine of (x squared) plus C.

Aligned with the College Board AP Calculus CED — Unit 6 (Integration and Accumulation of Change), Topic 6.9 (Integrating Using Substitution). College Board ↗

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