Integration Decision Guide: Which Technique?

Differentiation is an algorithm: apply the rules mechanically and you always finish. Integration is not. You are recognising a shape and matching it to a technique, and the failure mode that produces is thrashing — substitution, then parts, then substitution again with a different $u$, filling a page and finishing nowhere.

One piece of honesty before the tree, because it changes how you interpret being stuck. Not every elementary function has an elementary antiderivative. $e^{-x^2}$, $\sin (x)/x$, and $e^x/x$ are all perfectly well-behaved functions whose integrals provably cannot be written with the operations you know. On a set exam question this will not happen, so a wall you cannot get past is a wrong technique rather than an impossible integral. But the ceiling is why integration has techniques and a decision tree while differentiation has neither.

The most common wasted effort in a calculus exam is applying a technique to an integral that needed none. Rewrite every root and reciprocal as a power of $x$, split every sum, pull constants out. A surprising number of integrals surrender at that point.

The same branch swallows integrands that look like they need parts. Given $\int \frac{(x+1{)}^2}{x}dx$, the instinct is to see a quotient and reach for a rule. Expand the numerator and divide term by term instead:

Two lines, no technique. Algebra is free and techniques are expensive, so exhaust the algebra first.

Trigonometric integrands are where this pays best, because an identity often replaces an entire technique. There is no product rule to reverse for $\int {\sin }^{2}xdx$, but power reduction turns it into a sum: ${\sin }^{2}x=(1-\cos 2x)/2$, so the integral is $\frac{x}{2}-\frac{1}{4}\sin 2x+C$. Differentiating returns $\frac{1}{2}-\frac{1}{2}\cos 2x$, which is ${\sin }^{2}x$ again. The rule of thumb: an odd power of sine or cosine can be handled by substitution, because you can peel off one factor to serve as $du$, while an even power needs a power-reduction identity first. Recognising which identities are worth holding in memory is covered in the trig identities worth memorising.

Substitution is the chain rule read backwards. It works when the integrand contains a composite function and the derivative of the inner function, up to a constant multiple. Both halves are required, and checking the second takes about five seconds — which is why this is the cheapest branch to test.

On a definite integral, change the limits with the variable rather than substituting back at the end. For ${\int }_0^1{x}^2\sqrt{x^3+1}dx$, the limits $x=0$ and $x=1$ become $u=1$ and $u=2$, giving

Substitution also handles integrands that are not obviously composite. In $\int \frac{x}{x^2+4}dx$ the numerator is half the derivative of the denominator, so $u=x^2+4$ gives $\frac{1}{2}\ln (x^2+4)+C$. Any time a numerator is a constant multiple of the denominator's derivative, the answer is a logarithm and no other technique is needed. More on identifying the inner function, including the cases where two choices both work, is in the guide to u-substitution.

Occasionally the right $u$ is not the inside of a function but the whole awkward piece. For $\int \frac{x}{\sqrt{x+1}}dx$, setting $u=x+1$ gives $dx=du$ and, after rewriting the stray $x$ as $u-1$, the tidy integral $\int (u^{1/2}-u^{-1/2})du$. Setting $u=\sqrt{x+1}$ instead makes $x=u^2-1$ and $dx=2udu$, and the root cancels immediately, leaving $2\int (u^2-1)du$. Both close, and both give $\frac{2}{3}(x+1{)}^{3/2}-2\sqrt{x+1}+C$. Two viable choices is a comfortable position to be in; the useful reflex is to take whichever makes the remaining integrand a plain sum of powers.

Parts is the product rule read backwards: $\int udv=uv-\int vdu$. It trades one integral for another, so the only question that matters is whether the trade is favourable. LIATE answers it by ranking functions according to whether they improve more under differentiation than under integration.

Two variations are worth having seen. When the algebraic factor has degree two or more, parts has to be applied repeatedly, each round dropping the degree by one. For $\int x^2{e}^{x}dx$, the first round leaves $x^2{e}^x-2\int x{e}^{x}dx$, and the second finishes it:

Differentiating confirms it, since the product rule returns $e^x(x^2-2x+2)+e^x(2x-2)=x^2{e}^x$. The second variation is the circular case. With $I=\int e^x\sin xdx$, two rounds of parts return a multiple of $I$ itself, which looks like failure and is not — you solve for $I$ algebraically to get $I=\frac{1}{2}{e}^x(\sin x-\cos x)+C$. Both patterns, including how to recognise the circular case before you have gone round twice, are worked through in integration by parts and the LIATE rule.

There is a third case that looks like it cannot be a parts problem at all, because there is only one function. $\int \ln xdx$ has no visible product, and yet parts is the only way in. Take $u=\ln x$ and $dv=dx$ — the invisible factor of one becomes $dv$, so $v=x$. Then

Differentiating gives $\ln x+1-1=\ln x$, so it holds. The same manoeuvre integrates $\arctan x$ and $\arcsin x$, and it is the reason those two functions sit at the top of LIATE despite rarely appearing in a product: whenever one of them is the whole integrand, it still becomes $u$.

A quotient of polynomials always has an elementary antiderivative, which makes this the most reliable branch on the tree. It also has the most preliminaries: degrees first, then whether the denominator factors, and only then decompose.

When the numerator's degree is at least the denominator's, divide first. For $\int \frac{x^3}{x^2-1}dx$, division gives $x+\frac{x}{x^2-1}$, and the remainder term is a substitution with $u=x^2-1$:

When the denominator will not factor, partial fractions cannot start, and the move is to complete the square. The denominator of $\int \frac{dx}{x^2+6x+13}$ has discriminant $36-52=-16$, so it is irreducible; rewriting it as $(x+3{)}^2+4$ turns the integral into a standard arctangent form and gives $\frac{1}{2}\arctan \left(\frac{x+3}{2}\right)+C$. That is the same completing-the-square manipulation used to solve a quadratic equation, applied to a different end.

Repeated factors need one extra rule, and it is the rule most often got wrong. A factor appearing to the power $n$ contributes $n$ fractions, one for each power up to $n$, not a single fraction with the whole power in the denominator. So

Clearing gives $1=A(x+1{)}^2+Bx(x+1)+Cx$. Substituting $x=0$ gives $A=1$, and $x=-1$ gives $C=-1$. The repeated factor means substitution alone cannot find $B$, so compare coefficients of $x^2$: the left side has none, so $A+B=0$ and $B=-1$. A quick numerical check at $x=1$ gives one quarter on both sides. Integrating term by term produces $\ln |x|-\ln |x+1|+\frac{1}{x+1}+C$, and note that the squared factor integrates to a reciprocal rather than to a logarithm — mixing those two up is the other common error in this branch.

The most expensive technique here, and the one to reach for last. It applies when a root of a quadratic expression stands alone, with nothing outside it that could serve as $du$. It eliminates the root by invoking a Pythagorean identity, which is worth the overhead only because no cheaper technique can remove a root of that shape.

The tangent form works the same way. For $\int \frac{dx}{(x^2+4{)}^{3/2}}$, put $x=2\tan \theta$, so that $dx=2{\sec }^2\theta d\theta$ and the denominator becomes $(4{\sec }^2\theta {)}^{3/2}=8{\sec }^3\theta$. The integral reduces to $\frac{1}{4}\int \cos \theta d\theta$, and the triangle gives $\sin \theta =x/\sqrt{x^2+4}$, so the answer is $x/(4\sqrt{x^2+4})+C$. Notice how much cancellation the substitution produced: an integrand raised to the three-halves power became a single cosine.

On a definite integral, converting the limits skips the triangle entirely, and that saves more here than in ordinary substitution because the triangle is the slowest step. Bounds of $x=0$ to $x=3$ in the sine example become $\theta =0$ and $\theta =\pi /2$.

One caution about the third form. The secant substitution for $\sqrt{x^2-a^2}$ is genuinely harder than the other two, because $\sec \theta$ is not one-to-one over an interval containing both branches, so the sign of $\tan \theta$ depends on which side of the domain you are working. If a question hands you that shape, check first whether completing the square or a plain substitution avoids it. Frequently it does.

Choosing wrongly is normal and cheap, provided you notice quickly. The test is always structural: after two lines, is the integrand simpler — fewer radicals, lower degree, fewer kinds of function — or merely different? Different is a dead end. Here are the five you will actually meet.

A substitution that half works. Not every leftover $x$ is fatal. In $\int x^3\sqrt{x^2+1}dx$, the substitution $u=x^2+1$ absorbs one factor of $x$ into $du$ and leaves $x^2$, which you can rewrite as $u-1$. The integral becomes $\frac{1}{2}\int (u-1)\sqrt{u}du$ and closes cleanly at $\frac{1}{5}(x^2+1{)}^{5/2}-\frac{1}{3}(x^2+1{)}^{3/2}+C$. The distinction worth learning: a leftover even power of $x$ can always be converted, and an odd one cannot.

Parts chosen the wrong way round. On $\int x{e}^{x}dx$, taking $u=e^x$ and $dv=xdx$ gives $\frac{x^2}{2}{e}^x-\int \frac{x^2}{2}{e}^{x}dx$. The new integral has $x^2$ where the old one had $x$: the degree went up, so you are moving away from an answer. Swap the choices, which is what LIATE told you to do in the first place, and you get $x{e}^x-e^x+C$ in two lines.

Partial fractions on an irreducible denominator. Splitting $\int \frac{dx}{x^2+4}$ into two linear pieces fails immediately, because $x^2+4$ has discriminant $-16$ and so no real roots. Compute the discriminant before decomposing anything.

Partial fractions before long division. $\int \frac{x^2}{x^2+1}dx$ defeats decomposition because the fraction is improper. Divide first — $\frac{x^2}{x^2+1}=1-\frac{1}{x^2+1}$ — and the integral is $x-\arctan x+C$.

Trigonometric substitution when a plain one would do. $\int \frac{x}{\sqrt{x^2+9}}dx$ contains a root of the right shape, so the pattern-matching reflex says tangent. But there is an $x$ outside the root, so $u=x^2+9$ finishes it in one line at $\sqrt{x^2+9}+C$. Not a wrong answer, just an expensive route to a right one.

Integration is the rare topic where checking is strictly easier than solving, because differentiation is mechanical and integration is not. An antiderivative that does not differentiate back to the integrand is wrong, with no ambiguity and no partial credit to argue about, and the check takes a fraction of the time the integral did. Every worked example above ends with it for that reason.

The same asymmetry is why committing to a branch for two lines is cheap. If the integrand has not become structurally simpler by then, back out; you have spent thirty seconds rather than five minutes. If you would rather see the classification and the working side by side for your own integral, the integral calculator names the technique it chose before it starts, and the calculus solver handles the derivative check at the end. The two techniques you will use most are documented in depth in u-substitution: finding the inner function and integration by parts and the LIATE rule.