Techniques of Integration

Updated 7 October 2026

Differentiation follows rules that always work. Integration has no single procedure, so it is learned as a set of techniques, and the skill is reading the shape of the integrand and recognizing which technique it calls for.

What is integration?

Integration asks the reverse of differentiation: which function has this derivative? That function is an antiderivative, and every answer can be checked by differentiating it. For a power of x, add one to the power and divide by the new power, so $6x^2$6x² integrates to $2x^3+C$2x³ + C, because 6 ÷ 3 = 2. The rule undoes the power rule exactly: differentiating $2x^3$2x³ brings down the 3, which cancels the division, and lowers the power back to 2. A constant factor stays outside, and a sum is integrated term by term, so 4x + 3 gives $2x^2+3x+C$2x² + 3x + C, where + C covers any constant that differentiated away. Only $x^{-1}$x⁻¹ breaks the rule, since the new power would be 0; $\frac{1}{x}$1/x integrates to ln|x| + C instead. Differentiating $2x^3+C$2x³ + C gives back $6x^2$6x², since 3 × 2 = 6 and the C disappears. See The antiderivative.

$x^3$x³ integrates to $\frac{x^4}{4}$x⁴/4 by the same rule. See Integrating powers.

Why is there always a + C?

Differentiating removes any constant term, so integration cannot tell which constant was there, and + C stands for all of them. The functions $x^2$x², $x^2+7$x² + 7 and $x^2-100$x² − 100 all differentiate to 2x, so the integral of 2x is the family $x^2+C$x² + C, one parallel curve per value of C. Shifting a curve up or down never tilts it, so every member has the same gradient at the same x. One extra fact picks out a single curve: through (2, 9), 9 = 4 + C, so C = 5 and the function is $x^2+5$x² + 5. In a differential equation that fact is the initial condition. A definite integral needs no C, because it cancels in F(b) − F(a). Substituting x = 2 into $x^2+5$x² + 5 gives 4 + 5 = 9, the required height, and the gradient there is 2 × 2 = 4.

See Finding the constant of integration.

What does the integral sign actually say?

The sign $\int$∫ is a stretched S, for sum, and $\int f\left(x\right)dx$∫ f(x) dx means: add up thin slivers of height f(x) and width dx. The function f(x) is the integrand, and the numbers at the bottom and top are the limits of integration, where the adding starts and stops. The integral of 3 from x = 1 to x = 5 adds slivers of height 3 across a width of 4, since 5 − 1 = 4, so it totals 3 × 4 = 12, a rectangle's area. The notation records how the answer is built: height times width, summed, with the widths shrunk until the sum is exact. The dx names the variable being summed, which matters when other letters appear: integrating with dx holds any y constant. With no limits, the sign asks for an antiderivative plus a constant, not a number. The antiderivative 3x gives 15 − 3 = 12 as well. See Reading the integral sign.

dx has a meaning of its own. $\frac{dy}{dx}$dy/dx began as a rise over a run, and the pieces can be separated: if $y=x^2$y = x², then dy = 2x dx. Substitution depends on this. See Differentials.

Why does an integral measure area?

A definite integral is the limit of a sum of rectangles: slice the region into strips of width $\Delta x$Δx, treat each as a rectangle of height f(x), and add the areas $f\left(x\right)\Delta x$f(x) Δx. For y = x from 0 to 2, four strips of width $\frac{1}{2}$1/2 with right-edge heights give $\left(\frac{1}{2}+1+\frac{3}{2}+2\right)\times \frac{1}{2}=\frac{5}{2}$(1/2 + 1 + 3/2 + 2) × 1/2 = 5/2, over the true area, 2. The fundamental theorem skips the strips: the area from a to b is F(b) − F(a) for any antiderivative F, because the area up to x grows at the rate f(x). A total area needs one more step, since regions below the axis count as negative: split at each crossing and add the sizes. For y = x − 1 from 0 to 2, the integral is 0, but the area is $\frac{1}{2}+\frac{1}{2}=1$1/2 + 1/2 = 1. The triangle under y = x has area $\frac{1}{2}\times 2\times 2=2$1/2 × 2 × 2 = 2, matching $\frac{x^2}{2}$x²/2 at 2. See The definite integral.

Each height can be read at the left edge, the right edge or the middle of its strip. On a rising curve the left sum underestimates and the right sum overestimates. In sigma notation the sum is $\sum _{i=1}^{n}f\left(x_i\right)\Delta x$Σ f(xᵢ) Δx from i = 1 to n; let n run to infinity and $\sum$Σ becomes $\int$∫ and $\Delta x$Δx becomes dx. See Left, right and midpoint Riemann sums and The Riemann sum in sigma notation.

Let A(x) be the area under the curve up to x. Increase x a little and the area grows by a thin strip of height f(x), so A differentiates to f. See The fundamental theorem of calculus.

Area below the axis counts as negative. The arch of sin x from 0 to $\pi$π has integral 2, from $\pi$π to $2\pi$2π the integral is −2, and over the whole cycle it is 0. For the total area, split at the crossings and add the sizes: 2 + 2 = 4. See Area below the axis.

Swapping the limits changes the sign, a constant factor comes outside, and a to b plus b to c joins into a to c. See Properties of the definite integral.

How do you integrate a function inside another function?

Substitution is the chain rule read backwards: when an integral contains the derivative of an inside function as a factor, name the inside u, replace every x, and integrate in u. For the integral of $3x^2{\left(x^3+2\right)}^4$3x²(x³ + 2)⁴, let $u=x^3+2$u = x³ + 2, so $du=3x^{2}dx$du = 3x² dx and the integral becomes $\int u^{4}du=\frac{u^5}{5}+C$∫ u⁴ du = u⁵/5 + C, which is $\frac{{\left(x^3+2\right)}^5}{5}+C$(x³ + 2)⁵/5 + C. The factor $3x^2$3x² is absorbed exactly by du, leaving a plain power. A definite integral changes one step: convert the limits to u and never return to x. From x = 0 to 1, u runs from 2 to 3, so the integral is $\frac{\left(3^5-2^5\right)}{5}$(3⁵ − 2⁵)/5, and 243 − 32 = 211, giving $\frac{211}{5}$211/5. The right u is usually the expression inside a bracket or under a root, and it leaves no x behind. Differentiating $\frac{{\left(x^3+2\right)}^5}{5}$(x³ + 2)⁵/5 by the chain rule restores the factor $3x^2$3x². See Integration by substitution.

1. Choose u, usually the expression inside the bracket or under the root.
2. Find $\frac{du}{dx}$du/dx, and write dx in terms of du.
3. Replace every x. If one is left over, u was the wrong choice.
4. Integrate in u, then write the x expression back. For a definite integral, convert the limits instead.

$\int 2x{\left(x^2+1\right)}^{5}dx$∫ 2x(x² + 1)⁵ dx: 2x is the derivative of $x^2+1$x² + 1, so let $u=x^2+1$u = x² + 1 and du = 2x dx. The integral becomes $\int u^{5}du=\frac{u^6}{6}+C=\frac{{\left(x^2+1\right)}^6}{6}+C$∫ u⁵ du = u⁶/6 + C = (x² + 1)⁶/6 + C.

When the inside is linear, ax + b, its derivative is the constant a and there is a shortcut: integrate as if the inside were plain x, then divide by a. $\int \cos 3xdx=\left(\frac{1}{3}\right)\sin 3x+C$∫ cos 3x dx = (1/3) sin 3x + C. A curved inside has a derivative that changes with x, so the shortcut does not apply. See Integrating f(ax + b).

How do you integrate a product?

Integration by parts is the product rule read backwards: $\int udv=uv-\int vdu$∫ u dv = uv − ∫ v du. Choose u as the factor that simplifies when differentiated. For $\int x\cos xdx$∫ x cos x dx, let u = x and dv = cos x dx, so du = dx and v = sin x, and the integral is $x\sin x-\int \sin xdx$x sin x − ∫ sin x dx, which is x sin x + cos x + C. The formula is the product rule for uv, integrated and rearranged. The new integral must be easier, which is why a polynomial against $e^x$eˣ, sin x or cos x is u: differentiating lowers its power. A higher power needs more rounds, so $x^2{e}^x$x² eˣ takes two and gives $\left(x^2-2x+2\right)e^x+C$(x² − 2x + 2)eˣ + C. From 0 to $\pi$π the first answer gives (0 − 1) − (0 + 1) = −2, and differentiating x sin x + cos x returns x cos x. See Integration by parts.

$\int x{e}^{x}dx$∫ x eˣ dx: let u = x and $dv=e^{x}dx$dv = eˣ dx, so du = dx and $v=e^x$v = eˣ. Then $\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C$∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C. If the new integral is harder than the old one, swap the choice of u.

$\int \ln xdx$∫ ln x dx is done by parts with u = ln x and dv = 1 dx. $\int e^x\sin xdx$∫ eˣ sin x dx returns to itself after two rounds; treat the original integral as an unknown and solve for it.

How do you integrate trigonometric functions?

Integration reverses differentiation, so cos x integrates to sin x + C, and sin x to −cos x + C, since cos x differentiates to −sin x. Reversing other standard results gives ${\sec }^{2}x\to \tan x$sec²x → tan x and $\sec x\tan x\to \sec x$sec x tan x → sec x. With the angle in radians, the area under cos x from 0 to $\frac{\pi }{2}$π/2 is $\sin \left(\frac{\pi }{2}\right)-\sin 0$sin(π/2) − sin 0, which is 1 − 0 = 1. The minus sign on the cosine is the part most often lost. The results need radians because only there does sin x differentiate to exactly cos x; in degrees each would carry a factor of $\frac{180}{\pi }$180/π. Powers and products change the method. ${\sin }^{2}x$sin²x is not a standard derivative, so rewrite it with a double angle identity first, and sin x cos x becomes $\frac{\left(\sin 2x\right)}{2}$(sin 2x)/2, which integrates to $-\frac{\left(\cos 2x\right)}{4}+C$−(cos 2x)/4 + C. Differentiating −cos x + C returns sin x, the integrand, with the C gone.

${\sin }^{2}x$sin²x is not the derivative of any standard function, so rewrite it with the double angle formula: ${\sin }^{2}x=\frac{\left(1-\cos 2x\right)}{2}$sin²x = (1 − cos 2x)/2, and $\int {\sin }^{2}xdx=\frac{x}{2}-\frac{\left(\sin 2x\right)}{4}+C$∫ sin²x dx = x/2 − (sin 2x)/4 + C. ${\cos }^{2}x=\frac{\left(1+\cos 2x\right)}{2}$cos²x = (1 + cos 2x)/2 integrates the same way. The identities are in radians and trigonometric identities; see Integrating ${\sin }^2$sin² and ${\cos }^2$cos².

tan x is sin x over cos x, and the numerator is minus the derivative of the denominator, so $\int \tan xdx=-\ln |\cos x|+C=\ln |\sec x|+C$∫ tan x dx = −ln|cos x| + C = ln|sec x| + C. See Integrating ${\sec }^{2}x$sec²x, sec x tan x and tan x.

How do you integrate exponentials, logarithms and fractions?

$e^x$eˣ is its own derivative, so $\int e^{x}dx=e^x+C$∫ eˣ dx = eˣ + C, and $e^{2x}$e^(2x) integrates to $\frac{e^{2x}}{2}+C$e^(2x)/2 + C. Since ln x differentiates to $\frac{1}{x}$1/x, the integral of $\frac{1}{x}$1/x is ln|x| + C; the modulus keeps it defined for negative x. The logarithm fills the gap the power rule leaves, because raising the power −1 by one would mean dividing by 0. A fraction whose top is the derivative of its bottom integrates to the log of the bottom, since the chain rule makes $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$f'(x)/f(x) the derivative of ln f(x). So $\frac{3x^2}{\left(x^3+1\right)}$3x²/(x³ + 1) integrates to $\ln |x^3+1|+C$ln|x³ + 1| + C. A bottom that factors is split into partial fractions first. Stretching explains why ln 4 − ln 2 = ln 2: the strip under $\frac{1}{x}$1/x from 2 to 4 is the strip from 1 to 2 made twice as wide and half as tall, and $2\times \frac{1}{2}=1$2 × 1/2 = 1. See Integrating exponentials and logarithms.

Differentiating ln f(x) by the chain rule gives $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$f'(x)/f(x). Read backwards: when the numerator is the derivative of the denominator, the integral is ln of the denominator. $\int \frac{2x}{\left(x^2+1\right)}dx=\ln \left(x^2+1\right)+C$∫ 2x/(x² + 1) dx = ln(x² + 1) + C, because 2x is the derivative of $x^2+1$x² + 1. Differentiate the denominator first and check whether the numerator matches. See Integrals of f' over f.

When the denominator factors instead, split into partial fractions. Write $\frac{1}{\left(\left(x-1\right)\left(x+1\right)\right)}$1/((x − 1)(x + 1)) as $\frac{A}{\left(x-1\right)}+\frac{B}{\left(x+1\right)}$A/(x − 1) + B/(x + 1), clear the denominators, then substitute x = 1 to find A and x = −1 to find B. Each piece is a constant over a linear bracket, so each integrates to a logarithm. A repeated factor needs one fraction for each power; a quadratic that will not factor keeps a numerator Ax + B. See Integration by partial fractions.

Which integrals give inverse functions?

Two derivatives from the rules of differentiation, read in reverse, give inverse trigonometric answers. Since ${\tan }^{-1}x$tan⁻¹x differentiates to $\frac{1}{\left(1+x^2\right)}$1/(1 + x²), $\frac{1}{\left(1+x^2\right)}$1/(1 + x²) integrates to ${\tan }^{-1}x$tan⁻¹x, and since ${\sin }^{-1}x$sin⁻¹x differentiates to $\frac{1}{\sqrt{1-x^2}}$1/√(1 − x²), $\frac{1}{\sqrt{1-x^2}}$1/√(1 − x²) integrates to ${\sin }^{-1}x$sin⁻¹x. The area under $\frac{1}{\left(1+x^2\right)}$1/(1 + x²) from 0 to 1 is the angle whose tangent is 1, $\frac{\pi }{4}$π/4. With $a^2$a² in place of the 1, the answers become $\left(\frac{1}{a}\right){\tan }^{-1}\left(\frac{x}{a}\right)$(1/a) tan⁻¹(x/a) and ${\sin }^{-1}\left(\frac{x}{a}\right)$sin⁻¹(x/a), so $\frac{1}{\left(9+x^2\right)}$1/(9 + x²) gives $\left(\frac{1}{3}\right){\tan }^{-1}\left(\frac{x}{3}\right)+C$(1/3) tan⁻¹(x/3) + C, as 9 = 3 × 3. Substituting $x=\tan \theta$x = tan θ explains it: $1+x^2$1 + x² becomes ${\sec }^2\theta$sec²θ, which cancels $dx={\sec }^2\theta d\theta$dx = sec²θ dθ. Without the root, $\frac{1}{\left(1-x^2\right)}$1/(1 − x²) needs partial fractions instead. Differentiating $\left(\frac{1}{3}\right){\tan }^{-1}\left(\frac{x}{3}\right)$(1/3) tan⁻¹(x/3) gives two factors of $\frac{1}{3}$1/3, and $\left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)=\frac{1}{9}$(1/3) × (1/3) = 1/9 turns the result into $\frac{1}{\left(9+x^2\right)}$1/(9 + x²); see Integrals that give ${\tan }^{-1}$tan⁻¹ and ${\sin }^{-1}$sin⁻¹.

When $\sqrt{a^2-x^2}$√(a² − x²) appears with no factor of x beside it, substitute $x=a\sin \theta$x = a sin θ. Then $a^2-x^2=a^2{\cos }^2\theta$a² − x² = a² cos²θ, the root becomes $a\cos \theta$a cos θ, and $dx=a\cos \theta d\theta$dx = a cos θ dθ. A minus inside takes sine; a plus takes tangent. See Choosing a trigonometric substitution.

A plus under the root gives a hyperbolic answer: $\frac{1}{\sqrt{x^2+1}}$1/√(x² + 1) integrates to arsinh x, and $\frac{1}{\sqrt{x^2-1}}$1/√(x² − 1) to arcosh x. Both are logarithms, so either form is correct. See Integrals that give arsinh and arcosh.

How do you choose?

Read the shape of the integrand first and ask, in order: is it a standard result or a power? Is the top the derivative of the bottom, giving a logarithm? Is an inside function's derivative present, calling for substitution? Is it a product of unrelated families, such as x and $e^x$eˣ, calling for parts? Is it a fraction whose bottom factors, calling for partial fractions? Is it ${\sin }^{2}x$sin²x or ${\cos }^{2}x$cos²x, calling for a double angle identity? So $\frac{x}{\left(x^2+4\right)}$x/(x² + 4) passes the second test up to a factor of 2 and integrates to $\left(\frac{1}{2}\right)\ln \left(x^2+4\right)+C$(1/2) ln(x² + 4) + C. The early tests are quicker, and one integrand can pass several: $x{e}^{x^2}$x e^(x²) looks like a product, but substitution finishes it in a line. Some, such as $e^{-x^2}$e^(−x²), pass none, and a numerical method takes over. Differentiating the log gives 2x over the bottom, and $\frac{1}{2}\times 2=1$1/2 × 2 = 1 leaves $\frac{x}{\left(x^2+4\right)}$x/(x² + 4).

The mistakes worth naming

• Dropping the + C. An indefinite integral is a family of curves.
• Leaving an x behind after substituting. Every x must go, including the one in dx.
• Reporting a signed area as a total. Split at the crossings first.
• Choosing the wrong factor in parts. If the new integral is harder, swap.
• Forgetting the modulus in ln|x|. Without it the result fails for negative x.

The rules being reversed are in the rules of differentiation, their uses in applications of integration and polar curves, and the limits behind the definition in limits and continuity.