By This integral calculator finds antiderivatives with steps. Type a function of \( x \), press solve, and you get \( \int f(x)\,dx \) with the constant of integration, the rule that cracked it (power rule, substitution, integration by parts, standard trig and exponential forms) and a check that differentiating the answer gives back your integrand.
It works as an antiderivative calculator for homework, exam review, or just confirming a result you worked out by hand. The same solver has a Definite mode for integrals with limits. It is free and needs no sign-up.
How to use the integral calculator
- Type the integrand in the box. Use
^for powers,sqrt()for roots,e^(x)orexp()for exponentials,ln()for the natural log andsin(),cos(),sec()for trig functions. - Implicit multiplication works, so
3x,x^2 e^xandx ln(x)are read the way you would write them on paper. Use parentheses when in doubt, e.g.1/(x^2 + 4). - Press solve. The result appears with the rule used and a step list; you can copy it or plot it.
- For a number instead of a formula, switch to Definite. Then enter the lower and upper limits (
piandinfare allowed) and solve again.
How the calculator finds an antiderivative
An antiderivative of \( f \) is any function \( F \) with \( F'(x) = f(x) \). Every other antiderivative differs from it by a constant, which is why the answer ends in \( + C \):
$$\int f(x)\,dx = F(x) + C$$
The solver breaks sums into separate terms, pulls out constant factors, and then tries the standard tools in turn: the power rule \( \int x^n\,dx = \frac{x^{n+1}}{n+1} \) (with \( \ln|x| \) for \( n = -1 \)), known trig, exponential and inverse-trig forms, u-substitution when the derivative of an inner function is sitting in the integrand, and integration by parts for products like a polynomial times \( e^x \) or \( \ln x \). The last step differentiates the result to confirm it.
Worked example
Find \( \int x^2 e^x\,dx \). This is the example preloaded in the calculator above.
The integrand is a polynomial times an exponential, so use parts with \( u = x^2 \) and \( dv = e^x\,dx \). Then \( du = 2x\,dx \), \( v = e^x \), and
$$\int x^2 e^x\,dx = x^2 e^x - \int 2x\,e^x\,dx$$
The new integral needs parts once more (\( u = x \), \( dv = e^x\,dx \)), which gives \( \int x e^x\,dx = x e^x - e^x \). Put it together:
$$\begin{aligned}\int x^2 e^x\,dx &= x^2 e^x - 2(x e^x - e^x) + C \\ &= x^2 e^x - 2x e^x + 2e^x + C\end{aligned}$$
The calculator prints it in exactly this expanded form; factored, it is \( e^x(x^2 - 2x + 2) + C \). Check: differentiating gives \( 2xe^x + x^2e^x - 2e^x - 2xe^x + 2e^x = x^2 e^x \).
Want a number? In Definite mode from 0 to 1, the antiderivative is \( e \) at \( x = 1 \). It is also \( 2 \) at \( x = 0 \), so \( \int_0^1 x^2 e^x\,dx = e - 2 \approx 0.71828 \).
A quick substitution example to try next: type x/(x^2 + 9). With \( u = x^2 + 9 \), \( du = 2x\,dx \), the answer is \( \frac12 \ln(x^2 + 9) + C \).
What the solver can and cannot integrate
The step-by-step solver covers polynomials, roots and negative powers, \( e^{ax} \), sine and cosine of \( ax + b \), \( \sec^2 \), \( \tan \), \( \ln x \), \( \frac{1}{a^2 + x^2} \) forms, and products that yield to one substitution or repeated integration by parts.
Some integrands have no elementary antiderivative at all, such as \( e^{x^2} \) or \( \frac{\sin x}{x} \). Others, like \( \frac{1}{x^2 - 1} \), need partial fraction decomposition first. The solver does not do this automatically. In both cases it says that no closed form was found. You can switch to Definite mode and still get an accurate numerical value between any two limits.
Integration techniques at a glance
| Technique | Use it when | Example | ||
|---|---|---|---|---|
| Power rule | a sum of powers of \( x \) | \( \int x^3\,dx = \frac{x^4}{4} + C \) | ||
| Substitution | a function and its derivative appear | \( \int 2x\cos(x^2)\,dx = \sin(x^2) + C \) | ||
| Integration by parts | a product such as \( x e^x \) or \( x\ln x \) | \( \int xe^x\,dx = (x - 1)e^x + C \) | ||
| Partial fractions | a fraction of polynomials | \( \int \frac{dx}{x^2 - 1} = \frac{1}{2}\ln\left | \frac{x - 1}{x + 1}\right | + C \) |
| Trig identities | powers of sin and cos | \( \int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C \) |
See u-substitution, integration by parts and partial fraction decomposition.
Checking an antiderivative
Differentiate your answer: you should get back the integrand. For \( (x - 1)e^x \), the product rule gives \( e^x + (x - 1)e^x = xe^x \). This confirms the result.
Why the + C matters
Every function has infinitely many antiderivatives that differ by a constant, because the derivative of a constant is zero. In a definite integral the constant cancels, but in an indefinite integral, or when solving a differential equation, it must be included and then found from a starting condition.
Practice problems
- \( \int (6x^2 - 4x + 1)\,dx \)
- \( \int \frac{1}{x}\,dx \)
- \( \int x\ln x\,dx \)
Answers: 1. \( 2x^3 - 2x^2 + x + C \). 2. \( \ln|x| + C \). 3. By parts: \( \frac{x^2}{2}\ln x - \frac{x^2}{4} + C \).
Related guides
- Integration by Parts: Formula, LIATE Rule and Examples
- U-Substitution: Step-by-Step Method and Examples
- Fundamental Theorem of Calculus (Parts 1 and 2) Explained
- Partial Fraction Decomposition: Steps and Examples
Further reading
- Computing Indefinite Integrals (Paul’s Online Math Notes) — the basic integral formulas and how to rewrite an integrand before using them.
- Substitution Rule for Indefinite Integrals (Paul’s Online Math Notes) — how to choose \( u \), with many worked examples.
The full table of antiderivatives and techniques is in integration formulas. Following the Class 12 syllabus? See Integrals Class 12.
FAQ
Does the integral calculator show steps?
Yes. It lists the rule it applied, such as u-substitution with the choice of \( u \) and \( du \). Or it uses integration by parts with \( u \) and \( dv \). Then it adds \( + C \) and confirms the result by differentiating it.
Can it calculate definite integrals too?
Yes. Switch the solver to Definite, and enter the limits. The calculator evaluates \( F(b) - F(a) \) when it finds an antiderivative, or integrates numerically otherwise. Infinite limits such as inf are accepted. The definite integral calculator does the same and also says whether an improper integral converges.
Why is my answer different from the textbook’s?
Two correct antiderivatives can look different and still differ only by a constant. They might also be the same expression written another way (expanded versus factored). Differentiate both (the derivative calculator makes this quick); if you get the same integrand, both are right.
Is the antiderivative calculator free?
Yes. It is free to use with no sign-up and no limit on the number of integrals.