Integral calculator with steps

Type or scan an integral. HexMath integrates it and shows the substitution or parts at every step.

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What this page covers

Indefinite integrals of elementary functions, written with the constant + C.
Definite integrals with numeric limits, evaluated at the end.
Substitution, integration by parts, and standard forms for powers, exponentials, logarithms and trigonometric functions.
Not covered here: improper integrals with infinite limits, double and triple integrals, and numerical approximation of integrals with no closed form. You can still type one and see what the solver says.

Type ∫ from the math keyboard, or type "integral of" followed by the function and dx. Put limits on the integral sign for a definite integral.
Scan it. Fit the one integral in the frame.
Choose a photo. Several problems in one photo are solved as a list.
Then tap Solve.

Problem

\displaystyle \int x e^{2x}\,dx

Verified

The answer is

\dfrac{e^{2x}}{4}(2x - 1) + C

$Choose the parts$
$u = x$
$Differentiate and integrate the parts$
$du = dx$
$Apply integration by parts$
$\int u\,dv = uv - \int v\,du = \tfrac{x}{2}e^{2x} - \int \tfrac{1}{2}e^{2x}\,dx$
$Integrate the remaining term$
$\int \tfrac{1}{2}e^{2x}\,dx = \tfrac{1}{4}e^{2x}$
$Combine and add the constant$
$\tfrac{x}{2}e^{2x} - \tfrac{1}{4}e^{2x} + C$

Leaving out $+\,C$ on an indefinite integral.
Forgetting the $\tfrac{1}{2}$ when integrating $e^{2x}$ : the antiderivative is $\tfrac{1}{2}e^{2x}$ , not $e^{2x}$ .
Choosing $u = e^{2x}$ and $dv = x\,dx$ , which makes the remaining integral harder rather than simpler.
In a definite integral with substitution, keeping the old limits after changing the variable. Either change the limits with the substitution or substitute back before evaluating.

Check by differentiating. The derivative of $\frac{e^{2x}}{4}(2x - 1)$ is $x e^{2x}$ , the original integrand. If it is not, a step is wrong.
For a definite integral, check the number with a rough estimate: for the runnable example, the integrand is between 0 and 1 on $[0, 1]$ , so $\ln 2 \approx 0.69$ is plausible.
The integrand must be defined on the whole interval. On this page's examples it is; a denominator that reaches zero inside the limits makes the integral improper, which is outside this page.
HexMath can make mistakes. Double check important steps.

Yes. Write limits on the integral sign for a definite integral and you get a number; leave them off and you get an antiderivative with + C.

Yes. Each step has a short title, such as the substitution chosen or the parts used, followed by the working.

Yes. Use the camera on one problem, or choose a photo. Several problems in one photo are solved as a list.

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Last updated: 2026-09-22 · HexMath