You now have several techniques: substitution, parts, trig identities, trig substitution and partial fractions. In practice nobody tells you which one to use. Choosing is a skill of its own, and it improves with mixed practice.
Por qué funciona
Every technique runs a differentiation rule backward, so each one leaves a fingerprint. Substitution undoes the chain rule: look for an inside function whose derivative is also present. Parts undoes the product rule: look for a product of unlike functions. Partial fractions undoes adding fractions: look for a rational function. Trig substitution undoes the Pythagorean identities: look for $a^{2} \pm x^{2}$ or $x^{2} - a^{2}$ under a root.
Idea clave
Ask in this order: (1) Can algebra make it a basic integral? (2) Is part of it the derivative of another part (substitution)? (3) A rational function (divide, then partial fractions)? (4) A product of unlike functions (parts)? (5) Powers of trig functions (identities)? (6) A root of $a^{2} \pm x^{2}$ or $x^{2} - a^{2}$ (trig substitution)?
| If you see | Try first |
|---|---|
| $f(g(x))\,g'(x)$, like $x e^{x^{2}}$ or $\frac{\ln x}{x}$ | substitution |
| a power of $x$ times $e^{x}$, $\sin x$, $\ln x$ or $\arctan x$ | parts |
| a polynomial over a polynomial | division, then partial fractions |
| $\sin^{m}x\cos^{n}x$ or $\tan^{m}x\sec^{n}x$ | identities, then substitution |
| $\sqrt{a^{2} - x^{2}}$, $\sqrt{a^{2} + x^{2}}$, $\sqrt{x^{2} - a^{2}}$ with no spare $x$ | trig substitution |
Ejemplo resuelto: Three look-alikes
Find $\int \frac{x}{x^{2} + 4}dx$, $\int \frac{1}{x^{2} + 4}dx$ and $\int \frac{x^{2}}{x^{2} + 4}dx$.
- First: the top is half the derivative of the bottom. With $u = x^{2} + 4$: $\frac{1}{2}\ln(x^{2} + 4) + C$.
- Second: no spare $x$, but it matches the arctangent form: $\frac{1}{2}\arctan\frac{x}{2} + C$.
- Third: same degree on top, so divide: $\frac{x^{2}}{x^{2} + 4} = 1 - \frac{4}{x^{2} + 4}$, giving $x - 2\arctan\frac{x}{2} + C$.
Respuesta: Small changes in the integrand call for different methods.
Ejemplo resuelto: Two methods in a row
Find $\int e^{\sqrt{x}}\,dx$.
- No technique fits directly, so simplify the inside first: let $w = \sqrt{x}$, so $x = w^{2}$ and $dx = 2w\,dw$.
- The integral becomes $\int 2w e^{w}\,dw$, a product: use parts with $u = 2w$.
- $$\begin{aligned} &\int 2w e^{w}dw \\ &= 2w e^{w} - 2e^{w} + C \\ &= 2e^{\sqrt{x}}(\sqrt{x} - 1) + C. \end{aligned}$$
Respuesta: $2e^{\sqrt{x}}(\sqrt{x} - 1) + C$
Error común
Reaching for the heaviest tool first. For $\int \frac{x}{\sqrt{x^{2} + 4}}\,dx$, trig substitution works but takes a page; the substitution $u = x^{2} + 4$ gives $\sqrt{x^{2} + 4} + C$ in one line. Quick check before any technique: is there an inner function whose derivative (up to a constant) is sitting outside?
Some integrals have no antiderivative made of familiar functions at all: $\int e^{-x^{2}}dx$, $\int \frac{\sin x}{x}dx$, $\int \sqrt{1 + x^{3}}\,dx$. No technique will crack them. For their definite integrals, use numerical integration or series.