U Substitution Step by Step (With Examples)

The chain rule says $\frac{d}{dx}\left[F(g(x))\right] = F'(g(x)) \cdot g'(x)$. Reading that backward gives the substitution rule:

$$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du \quad \text{where } u = g(x)$$

So u substitution works when the integral contains an inside function and that inside function's derivative (up to a constant factor).

How to choose u. Good choices are usually:

• The expression inside parentheses raised to a power, like $x^2 + 1$ in $(x^2 + 1)^5$.
• The expression under a square root.
• The exponent in $e^{\text{something}}$.
• The input of a trig function, like $x^2$ in $\cos(x^2)$.
• A denominator whose derivative appears in the numerator.

1. Choose $u$, usually the inside function.
2. Find $du$ by differentiating: $du = u'\,dx$.
3. Rewrite the whole integral in terms of $u$. No $x$ and no $dx$ may be left over.
4. Integrate with respect to $u$.
5. Substitute back to get an answer in terms of $x$, and add $+ C$.

Example: Find $\int 2x(x^2 + 1)^5\,dx$.

1. Let $u = x^2 + 1$.
2. Then $du = 2x\,dx$.
3. The $2x\,dx$ in the integral is exactly $du$, so the integral becomes $\int u^5\,du$.
4. Integrate: $\frac{u^6}{6} + C$.
5. Substitute back: $\frac{(x^2 + 1)^6}{6} + C$.

Check by differentiating: $\frac{6(x^2 + 1)^5 \cdot 2x}{6} = 2x(x^2 + 1)^5$. It matches the integrand.

Often $du$ is off by a constant factor. Solve for the piece you have and pull the constant out front.

Example 1: Find $\int x\cos(x^2)\,dx$.

1. Let $u = x^2$, so $du = 2x\,dx$.
2. The integral only has $x\,dx$, so divide by 2: $x\,dx = \frac{1}{2}du$.
3. Rewrite: $\int \cos u \cdot \frac{1}{2}\,du = \frac{1}{2}\int \cos u\,du$.
4. Integrate: $\frac{1}{2}\sin u + C$.
5. Substitute back: $\frac{1}{2}\sin(x^2) + C$.

Example 2: Find $\int \frac{x^2}{x^3 + 4}\,dx$.

1. Let $u = x^3 + 4$, so $du = 3x^2\,dx$ and $x^2\,dx = \frac{1}{3}du$.
2. Rewrite: $\frac{1}{3}\int \frac{1}{u}\,du$.
3. Integrate and substitute back: $\frac{1}{3}\ln|x^3 + 4| + C$.

You can only fix constants this way. If you are missing a factor of $x$, you cannot pull it outside the integral.

With a definite integral, the bounds are $x$ values. Once you switch to $u$, convert them to $u$ values using your substitution. Then you never have to switch back.

Example 1: Evaluate $\int_0^2 x(x^2 + 1)^3\,dx$.

1. Let $u = x^2 + 1$, so $du = 2x\,dx$ and $x\,dx = \frac{1}{2}du$.
2. Change the bounds: when $x = 0$, $u = 0^2 + 1 = 1$. When $x = 2$, $u = 2^2 + 1 = 5$.
3. Rewrite: $\frac{1}{2}\int_1^5 u^3\,du$.
4. Integrate: $\frac{1}{2} \cdot \frac{u^4}{4}\Big|_1^5 = \frac{u^4}{8}\Big|_1^5$.
5. Evaluate: $\frac{5^4 - 1^4}{8} = \frac{625 - 1}{8} = \frac{624}{8} = 78$.

Example 2: Evaluate $\int_1^e \frac{\ln x}{x}\,dx$.

1. Let $u = \ln x$, so $du = \frac{1}{x}\,dx$.
2. Change the bounds: $x = 1$ gives $u = \ln 1 = 0$. $x = e$ gives $u = \ln e = 1$.
3. Rewrite: $\int_0^1 u\,du$.
4. Evaluate: $\frac{u^2}{2}\Big|_0^1 = \frac{1}{2} - 0 = \frac{1}{2}$.

If you would rather keep the original bounds, substitute back to $x$ first and then plug in the $x$ bounds. For Example 1 that gives $\frac{(x^2 + 1)^4}{8}\Big|_0^2 = \frac{625 - 1}{8} = 78$, the same answer. Just never mix $u$ expressions with $x$ bounds.

Sometimes an $x$ remains after substituting. If so, solve your $u$ equation for $x$ and replace it.

Example: Find $\int x\sqrt{x - 1}\,dx$.

1. Let $u = x - 1$, so $du = dx$. That leaves the outside $x$ unaccounted for.
2. Solve for $x$: $x = u + 1$.
3. Rewrite: $\int (u + 1)u^{1/2}\,du = \int \left(u^{3/2} + u^{1/2}\right)du$.
4. Integrate with the power rule: $\frac{2}{5}u^{5/2} + \frac{2}{3}u^{3/2} + C$.
5. Substitute back: $\frac{2}{5}(x - 1)^{5/2} + \frac{2}{3}(x - 1)^{3/2} + C$.

If no choice of $u$ makes the integral simpler, the problem probably needs another technique, such as integration by parts. Paste it into the VeilGPT solver to check your steps.