Inner function x²
- Input
- ∫ 2x·cos(x^2) dx
- Expected output
- sen(x²) + C
u = x², du = 2x dx.
u-substitution integral calculator
Substitution undoes the chain rule. When the integrand has the form f(g(x))·g′(x), replacing g(x) with u turns the integral into ∫ f(u) du. The tool tests the inner functions of the integrand as candidates, checks that the leftover x disappears and shows u, du and the way back.
u = x², du = 2x dx.
u = x² + 1.
u = 1 + eˣ, du = eˣ dx.
u = ln x, du = dx/x.
u = 3x + 2, du = 3 dx.
The indefinite integral of f is the family of functions F whose derivative is f. Since the derivative of a constant is zero, if F is an antiderivative then F + C is too, for any number C. That is why ∫ 2x dx = x² + C: x² + 1 and x² − 7 also have derivative 2x. The constant can only be pinned down with an extra condition, such as F(0) = 3.
Then the method may be another one: parts for products such as x·eˣ, partial fractions for quotients of polynomials, or trigonometric identities for powers of sin and cos. The calculator goes through all these rules on its own, in the order in which they usually solve fastest.
Yes when the argument can be negative: ∫ du/u = ln|u| + C. The tool drops the absolute value when the argument is always positive, as in ln(1 + eˣ) or ln(x² + 1).
Not literally. Two antiderivatives differ by a constant, and the same function can be written in different forms. Differentiate the answer key: if you get the integrand, it is equivalent.
Differentiating the result
F′(x) = f(x) at 14 test points (maximum error 2.2e-16). Both expressions are also identical after expanding.
| x | f(x) | F′(x) | Difference |
|---|---|---|---|
| -2.26376 | -0.23533751 | -0.23533751 | 2.8e-17 |
| 0.565523 | 0.99552794 | 0.99552794 | 1.1e-16 |
| 4.44149 | 377.08727 | 377.08727 | 5.7e-14 |
| -0.222645 | -0.17820488 | -0.17820488 | 0 |
| -0.357476 | -0.25003303 | -0.25003303 | 0 |
| 0.460526 | 0.72989202 | 0.72989202 | 2.2e-16 |
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