U-Substitution Examples: Choose u and Check Each Integral

Consider $\int 2x\cos(x^2)\,dx$. No basic rule integrates a product like that directly, yet the answer is short: $\sin(x^2)+C$. Differentiate $\sin(x^2)$ with the chain rule and you get $\cos(x^2)\cdot2x$, the integrand. U-substitution is the method that finds such answers on purpose. It runs the chain rule backwards: spot an inner function, rename it $u$, and let its derivative absorb the leftover factor.

This guide works through six u-substitution examples, from the plain case to constant factors, logarithms, definite integrals with new limits, and a substitution that needs a back-solve for $x$. If the chain rule itself still feels shaky, warm up with our chain rule worked examples first; every check below depends on it.

The rule in one line

OpenStax Calculus, Volume 1, §5.5 states substitution for indefinite integrals: if $u=g(x)$ and $F$ is an antiderivative of $f$, then

$$\int f(g(x))\,g'(x)\,dx=F(g(x))+C.$$

In practice you write $u=g(x)$ and $du=g'(x)\,dx$, rewrite the whole integral in $u$, integrate, and substitute back. The test of a good choice is simple: after the rewrite, no $x$ may remain. If one does, either solve for it in terms of $u$ or choose a different $u$.

How to choose u

Look for a function sitting inside another function: inside a power, under a root, in an exponent, or inside a trigonometric or logarithmic function. Then ask whether its derivative, possibly times a constant, multiplies the rest.

The last row matters. $\int e^{x^2}\,dx$ looks similar to the second row, but without the factor $x$ there is nothing for $du$ to absorb, and that integral has no elementary antiderivative at all. Substitution is not a trick that always works; it works when the derivative is there.

Example 1: the derivative is already present

Find $\int 2x(x^2+1)^5\,dx$. Let $u=x^2+1$, so $du=2x\,dx$. The factor $2x\,dx$ is exactly $du$:

$$\int 2x(x^2+1)^5\,dx=\int u^5\,du=\frac{u^6}{6}+C=\frac{(x^2+1)^6}{6}+C.$$

Check: the derivative of $\frac{(x^2+1)^6}{6}$ is $(x^2+1)^5\cdot2x$. Expanding $(x^2+1)^5$ and integrating term by term would also work, but it takes far longer and invites arithmetic slips.

Example 2: a constant factor is missing

Find $\int xe^{x^2}\,dx$. With $u=x^2$ we get $du=2x\,dx$, but the integrand has only $x\,dx$. Divide by two: $x\,dx=\frac{1}{2}\,du$.

$$\int xe^{x^2}\,dx=\frac{1}{2}\int e^{u}\,du=\frac{1}{2}e^{x^2}+C.$$

Moving $\frac{1}{2}$ outside is allowed because it is a constant. You may not do the same with a missing variable factor: $\int e^{x^2}\,dx$ cannot be rescued by writing $dx=\frac{du}{2x}$, because that leaves $x$ inside the $u$-integral.

Example 3: a logarithm as the inner function

Find $\int\frac{\ln x}{x}\,dx$ for $x>0$. Let $u=\ln x$, so $du=\frac{1}{x}\,dx$:

$$\int\frac{\ln x}{x}\,dx=\int u\,du=\frac{u^2}{2}+C=\frac{(\ln x)^2}{2}+C.$$

A related standard result comes from $\int\tan x\,dx=\int\frac{\sin x}{\cos x}\,dx$. Let $u=\cos x$, so $du=-\sin x\,dx$ and the integral becomes $-\int\frac{du}{u}=-\ln\lvert\cos x\rvert+C$. Watch the minus sign: it comes from the derivative of cosine, and dropping it is the usual error in this example.

Example 4: a definite integral with new limits

Evaluate $\int_0^1 x(1+x^2)^3\,dx$. Let $u=1+x^2$, so $du=2x\,dx$ and $x\,dx=\frac{1}{2}\,du$. Now convert the limits: when $x=0$, $u=1$; when $x=1$, $u=2$.

$$\int_0^1 x(1+x^2)^3\,dx=\frac{1}{2}\int_1^2 u^3\,du=\frac{1}{2}\left[\frac{u^4}{4}\right]_1^2=\frac{1}{2}\cdot\frac{16-1}{4}=\frac{15}{8}.$$

Once the limits are in $u$, there is no need to substitute back. OpenStax gives this as the definite-integral form of the rule: the limits change to $g(a)$ and $g(b)$. The classic mistake is to use the new integrand with the old limits 0 and 1, which here would give $\frac{1}{8}$, an answer off by a factor of 15.

Sanity check: the integrand rises from 0 at $x=0$ to 8 at $x=1$, so the area should be somewhere between 0 and 8, and likely nearer the low end because the curve is steep near the right edge. $\frac{15}{8}=1.875$ fits.

Example 5: solve for the leftover x

Find $\int x\sqrt{x+1}\,dx$. Take $u=x+1$, so $du=dx$. One $x$ remains outside the root. Solve the substitution for it: $x=u-1$.

$$\int x\sqrt{x+1}\,dx=\int(u-1)u^{1/2}\,du=\int\left(u^{3/2}-u^{1/2}\right)du=\frac{2}{5}(x+1)^{5/2}-\frac{2}{3}(x+1)^{3/2}+C.$$

This back-solve is legitimate because $u=x+1$ is easy to invert. When a remaining $x$ cannot be expressed in $u$ neatly, the substitution is probably the wrong one.

Example 6: when substitution is the wrong tool

Try $\int xe^{x}\,dx$. The obvious candidate $u=x$ changes nothing, and $u=e^x$ leaves $x=\ln u$ inside, turning the problem into $\int\ln u\,du$, which needs another method anyway. This product has no inner function with its derivative alongside. It is a case for integration by parts, where one factor gets simpler when differentiated.

A useful habit: spend at most a minute testing one or two choices of $u$. If none makes the $x$ disappear, switch tools rather than forcing it.

Mistakes to avoid

• Forgetting du. Writing $u=x^2$ and then replacing only the inner function leaves $dx$ in an integral over $u$. Every $dx$ must become a multiple of $du$.
• Dividing by a variable. Only constant factors can be moved outside the integral.
• Old limits on a new integral. Convert both limits, or go back to $x$ before evaluating.
• Leaving the answer in u. For an indefinite integral, substitute back so the answer is a function of $x$, and keep the $+C$.
• Skipping the check. Differentiating the answer takes seconds and catches nearly every slip with signs and constants.

Practice problems with answers

1. $\int 3x^2(x^3-4)^4\,dx$. Let $u=x^3-4$. Answer: $\frac{(x^3-4)^5}{5}+C$.
2. $\int\sin^2x\cos x\,dx$. Let $u=\sin x$. Answer: $\frac{\sin^3x}{3}+C$.
3. $\int e^{3x}\,dx$. Let $u=3x$, so $dx=\frac{1}{3}\,du$. Answer: $\frac{e^{3x}}{3}+C$.
4. $\int_1^{e}\frac{\ln x}{x}\,dx$. Let $u=\ln x$, with limits 0 and 1. Answer: $\frac{1}{2}$.

Substitution is the first technique to try on almost any integral, which is why our integrals guide calls it the workhorse of the chapter. Name the inner function, check that its derivative is there, convert everything including the limits, and differentiate the result before moving on.