Integration Techniques

Calculus is the math of change. One half measures how fast something is changing at a single instant, and the other half adds up small amounts to give a total.

Integration Techniques: the central idea

Integration techniques reverse familiar differentiation patterns. The structure of the integrand—not a memorized keyword—determines which technique reduces the work.

Words you need

Antiderivative

An antiderivative of a function is any function whose derivative gives you that function back, which is why every antiderivative comes with a $+C$.

Integrand

The integrand is the expression sitting between the integral sign and the $dx$, that is, the thing you are trying to undo.

Substitution

Substitution is a technique that renames an inner expression as $u$ so the integral turns into a simpler one, and it works because it is the chain rule read in reverse.

Integration by parts

Integration by parts is a technique that trades one integral for an easier one using $\int u\,dv = uv-\int v\,du$, and it comes from reversing the product rule.

Partial fractions

Partial fractions is a technique that breaks one complicated fraction into a sum of small fractions, each with a simple factor on the bottom, so each piece integrates on its own.

Constant of integration

The constant of integration, written $+C$, records that shifting a function up or down never changes its slope, so infinitely many antiderivatives share the same derivative.

What to know before this lesson

Review antiderivative rules, the chain and product rules, algebraic simplification, and definite-integral bounds.

If one of those prerequisites is uncertain, use the Calculus subject guide to locate the earlier concept before memorizing a procedure.

Integration techniques: a worked example

Every step, with the arithmetic

1. Step 1 - Look at the integrand$\int 6x^2(x^3+4)^5\,dx$ contains $x^3+4$ and, nearby, something proportional to its derivative
2. Step 2 - Name the inner functionLet $u=x^3+4$
3. Step 3 - Differentiate the substitution$du=3x^2\,dx$, so $x^2\,dx=du/3$
4. Step 4 - Rewrite every piece in terms of $u$$\int 6x^2(x^3+4)^5\,dx=\int 6u^5\cdot\frac{du}{3}=\int 2u^5\,du$
5. Step 5 - Integrate the simple version$2u^6/6=u^6/3$
6. Step 6 - Go back to $x$ and add the constant$(x^3+4)^6/3 + C$
7. Step 7 - Check by differentiating$\frac{6(x^3+4)^5\cdot3x^2}{3}=6x^2(x^3+4)^5$, the original integrand
8. Step 8 - Same integral with bounds 0 to 1$(1^3+4)^6/3-(0^3+4)^6/3=(15625-4096)/3=11529/3=3843$

In $\int2x\cos(x^2)dx$, choosing $u=x^2$ makes $du=2x\,dx$. The integral becomes $\int\cos u\,du=\sin u+C$.

Which technique to reach for, based on what the integrand looks like

The step-by-step method for integration techniques

1. Simplify the integrand and look for an inner function with its derivative, a product, a rational expression, or a trigonometric identity.
2. Choose substitution, integration by parts, partial fractions, or another technique and transform every factor including $dx$.
3. Integrate, return to the original variable, add $C$ for an indefinite integral, and differentiate to verify.

How to check your answer

Differentiate $\sin(x^2)+C$ to obtain $2x\cos(x^2)$, exactly the original integrand.

Compare units and qualitative behavior. A derivative should match the graph's slope, while a definite integral should agree with the signed area and scale of the interval.

A mistake that changes the mathematics

A substitution is incomplete if part of the old variable remains. After changing to $u$, the entire integrand and differential must use $u$.

Where you will actually use this

Total distance from speed

If a drone flies at $v(t)=6t$ metres per second, integrating from $t=0$ to $t=5$ gives $3t^2$ evaluated over that range, or $75-0=75$ metres travelled.

Volume of an odd-shaped tank

Engineers slice a curved tank into thin discs, write the area of one disc as a function of height, and integrate to get the total capacity.

Work done by a stretching spring

Force changes as a spring stretches, so physicists integrate force over distance rather than multiplying, which is what turns $F=kx$ into work $=\tfrac12kx^2$.

How integration techniques connects to the rest of calculus

• Fundamental theorem of calculus — Once you have an antiderivative, that theorem is what lets you turn it into an exact area or total.
• Derivative rules — Substitution reverses the chain rule and integration by parts reverses the product rule, so every technique here is a derivative rule read backwards.
• Factoring polynomials — Partial fractions cannot start until the bottom of the fraction has been factored.

Try a transfer problem

Evaluate $\int x e^x dx$ by parts and explain why choosing $u=x$ makes the remaining integral simpler.

Show the worked answer

For $\int x e^x\,dx$, choose $u=x$ and $dv=e^x dx$. Then $du=dx$ and $v=e^x$. The formula $\int u\,dv = uv-\int v\,du$ gives $\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C$, which factors as $e^x(x-1)+C$. Choosing $u=x$ is what makes the leftover integral simpler: differentiating $x$ turns it into $1$, so the polynomial disappears and only $\int e^x dx$ is left. The other choice, $u=e^x$ and $dv=x\,dx$, gives $v=x^2/2$ and leaves $\int \tfrac{x^2}{2}e^x dx$, which has a higher power of $x$ than you started with, so the problem got worse. The rule of thumb is to let $u$ be the factor that gets simpler when differentiated. Check the answer: $\frac{d}{dx}[xe^x-e^x] = (e^x + xe^x) - e^x = xe^x$.

Work without copying the example. When finished, use the relevant focused calculator or formula reference to check the setup and result, then correct the first line where your reasoning changed. When the method feels reliable, move to calculus practice questions.

Questions about integration techniques

How do I choose $u$ for integration by parts?

Use the LIATE order: Logarithm, Inverse trig, Algebraic (polynomials), Trig, Exponential. Whichever type appears first in that list becomes $u$. In $\int x\ln x\,dx$ the logarithm wins, so $u=\ln x$, and the answer is $\tfrac{x^2}{2}\ln x-\tfrac{x^2}{4}+C$.

Do I have to change the limits when I substitute in a definite integral?

You have a choice. Either convert the limits along with the variable, so $x$ from $0$ to $1$ with $u=x^3+4$ becomes $u$ from $4$ to $5$, or finish in $u$, switch back to $x$, and then plug in the original limits. Mixing them up, such as putting $x$-limits into a $u$-expression, is the classic error.

Why does $\int 1/x\,dx$ give $\ln|x|$ instead of following the power rule?

The power rule would ask you to raise the power from $-1$ to $0$ and divide by $0$, which is not allowed. The function $\ln|x|$ fills that one gap because its derivative is exactly $1/x$, and the absolute value bars let it work for negative $x$ too.

Does every function have an antiderivative you can write down?

No. Every continuous function has an antiderivative, but some cannot be written with the usual functions. $\int e^{-x^2}dx$ is the famous one behind the normal distribution; it is real and well defined, yet it has no formula in terms of powers, roots, logs, and trig, so it is computed numerically.