Calculator

Integral Calculator with Steps

Evaluate definite and indefinite integrals with the technique named before it is used, every substitution written in full, and the limits converted rather than quietly carried along.

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How to use it

Write the integrand and say whether there are limits: “integrate x cos x dx” and “evaluate the integral of x cos x from 0 to pi” produce different work. Name the variable if it is not $x$ . If you already tried a substitution and got stuck, paste your attempt in — the follow-up questions on a solution are free, so it is cheaper to ask why your $u$ failed than to start over. The calculus solver takes the same input when the integral is one part of a longer question.

The method behind it

Integration is differentiation run backwards. An antiderivative of $f$ is any function whose derivative is $f$ , and because constants differentiate to zero there is always a whole family of them — hence the $+ C$ .

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C (n \neq = - 1), \int \frac{1}{x} d x = ln ∣ x ∣ + C

A definite integral is a number, not a family, and the fundamental theorem of calculus is what connects the two ideas. It says the signed area under a curve is the change in any antiderivative across the interval, so the $C$ cancels.

\int_{a}^{b} f (x) d x = F (b) - F (a) where F^{'} = f

Only two techniques do most of the work. Substitution reverses the chain rule: if the integrand contains a composite function $f (g (x))$ multiplied by $g^{'} (x)$ , then setting $u = g (x)$ converts the whole thing into a standard form.

\int f (g (x)) g^{'} (x) d x = \int f (u) d u

Integration by parts reverses the product rule. Integrating both sides of $(uv)^{'} = u^{'} v + u v^{'}$ and rearranging gives the formula, and the whole game is choosing $u$ so that the integral on the right is easier than the one you started with.

\int u d v = uv - \int v d u

LIATE is the ordering that usually gets that choice right: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. Whichever factor appears earlier in that list becomes $u$ , because differentiating it simplifies the problem while integrating it would not. The reasoning behind the ordering is set out in the integration by parts and LIATE guide, and choosing the inner function for a substitution covers the other half.

One structural point worth internalising: substitution needs the inner derivative to already be present, by parts does not. If you scan for the derivative and it is genuinely absent, stop looking and try by parts, partial fractions, or an algebraic rewrite instead of forcing a $u$ that will not close.

Worked examples

Worked example

\int x cos x d x

1
A polynomial multiplied by a trig function, and the derivative of x is not lurking anywhere useful. LIATE puts algebraic ahead of trigonometric, so x becomes u.
$u = x, d v = cos x d x$
2
Differentiate u and integrate dv. The constant of integration is left out here; it reappears at the end.
$d u = d x, v = sin x$
3
Substitute into the by-parts formula.
$\int x cos x d x = x sin x - \int sin x d x$
4
The remaining integral is standard, and note the double negative: the antiderivative of sine is minus cosine.
$x sin x - (- cos x) + C$
5
Check by differentiating: the product rule gives sin x + x cos x, and the last term contributes minus sin x, leaving exactly x cos x.
$\frac{d}{d x} (x sin x + cos x) = sin x + x cos x - sin x = x cos x$

Answer

\int x cos x d x = x sin x + cos x + C

Worked example

\int_{0}^{2} x e^{x^{2}} d x

1
The exponent is a composite and its derivative, 2x, is present up to the constant 2. That is the signal for substitution.
$u = x^{2}, d u = 2 x d x \Rightarrow x d x = \frac{1}{2} d u$
2
Convert the limits now rather than back-substituting later. They are x-values, and the new integral is in u.
$x = 0 \Rightarrow u = 0, x = 2 \Rightarrow u = 4$
3
Rewrite the whole integral in u. Nothing in x may survive this step.
$\int_{0}^{2} x e^{x^{2}} d x = \frac{1}{2} \int_{0}^{4} e^{u} d u$
4
Integrate and apply the new limits.
$\frac{1}{2} [e^{u}]_{0}^{4} = \frac{1}{2} (e^{4} - e^{0})$
5
Simplify. To three decimal places this is 26.799, which is a useful sanity check against a numerical estimate.
$\frac{e ^{4} - 1}{2}$

Answer

\int_{0}^{2} x e^{x^{2}} d x = \frac{e ^{4} - 1}{2} \approx 26.799

Which technique should you use?

Most wasted time on an integral is spent committing to the wrong technique and pushing on. Two questions — is there a composite with its own derivative present, and are the factors of genuinely different types — settle the choice before you write a line.

Choose an integration technique

Look at the integrand as a whole. Which description fits it?

Common mistakes

Losing the constant of integration

An indefinite integral is a family of functions, not one function. Write $+ C$ as you write the antiderivative, not as an afterthought. In differential equations the missing constant is not a presentation issue — it destroys the solution.

Leaving a stray x after substituting

Once you commit to $u$ , every $x$ and the $d x$ must be converted. If an $x$ survives, the substitution has not worked; either solve for $x$ in terms of $u$ and continue, or abandon that choice of $u$ .

Keeping the old limits

In a definite integral the limits are values of $x$ . After substituting they must become values of $u$ , or you must back-substitute to $x$ before evaluating. In the example above, using 0 to 2 on the $u$ integral would give $(e^{2} - 1) /2 \approx 3.19$ instead of 26.80.

Practice

Answers are checked here — nothing is sent anywhere.

1
$\int (6 x^{2} - 4 x) d x$
2
$\int_{1}^{3} 2 x d x$
3
$\int ln x d x$

Frequently asked questions

Does it handle definite and indefinite integrals?

Both. Give limits and you get the antiderivative, the substitution of each limit, and the subtraction written out. Leave them off and you get the general antiderivative with its constant of integration.

Will it show the substitution, or just the answer?

The choice of u, the computation of du, the rewritten integral in terms of u, and the back-substitution are separate steps. That is where substitution problems actually go wrong, so those lines are the point.

Why can some integrals not be done at all?

Some integrands have no antiderivative expressible with the usual functions. The bell curve integrand e^(−x²) is the standard example. No calculator can produce one because it does not exist in closed form, though a definite integral of it can still be evaluated numerically.

Can it check an antiderivative I already have?

Yes, and you can also check it yourself in ten seconds: differentiate your answer. If it does not return the original integrand exactly, something is wrong, and the derivative usually shows you which factor you dropped.

What does it cost?

The first solution is free with no account. A free account allows three solutions a day; Gauth Plus is unlimited at $11.99 a month with a three-day free trial.

Keep going

Calculus solverIntegrals alongside derivatives, limits, and series in a single worked solution.U-substitution: finding the inner functionHow to spot when substitution will close, and how to change the limits.Integration by parts and LIATEChoosing u correctly, repeating the rule, and recognising the circular case.Derivative calculatorDifferentiate your antiderivative to confirm it returns the integrand.