Example 1, Power rule for integration
∫ x² dx
x³/3 + C
- 1
Apply power rule
n = 2, so add 1 to exponent and divide by new exponent - 2
Result
x³ / 3 + C
Power rule: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ −1.
Power · u-sub · By Parts
Integral Calculator ∫ Antiderivatives
Compute definite and indefinite integrals using the power rule, substitution, and integration by parts. Five worked examples cover the main techniques.
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From derivatives back to functions
Two Foundational Identities
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C · ∫ u dv = uv − ∫ v du
Power rule for polynomials, integration by parts for products. With u-substitution, these three techniques cover most freshman-calculus integrals.
Integration is the inverse of differentiation. The indefinite integral ∫ f(x) dx gives the family of all antiderivatives of f. The constant of integration + C represents that ambiguity, every antiderivative differs from another by a constant.
The definite integral ∫ₐᵇ f(x) dx gives a single number, the signed area under the curve from a to b. The Fundamental Theorem of Calculus connects the two: evaluate any antiderivative at the bounds and subtract.
Worked Examples
Five integrals across four techniques
Power rule, sums, definite integral via FTC, u-substitution, and integration by parts.
Example 2, Sum and constant multiple
∫ (3x² + 4x − 5) dx
x³ + 2x² − 5x + C
- 1
Split into terms
∫ 3x² dx + ∫ 4x dx − ∫ 5 dx - 2
Apply power rule to each
3·(x³/3) + 4·(x²/2) − 5x - 3
Simplify
x³ + 2x² − 5x + C
Integrals distribute over addition.
Example 3, Definite integral
∫₀² x² dx
8/3
- 1
Find antiderivative
F(x) = x³/3 - 2
Evaluate at bounds
F(2) − F(0) = 8/3 − 0 - 3
Result
8/3
Fundamental theorem of calculus: F(b) − F(a).
Example 4, Substitution (u-sub)
∫ 2x·cos(x²) dx
sin(x²) + C
- 1
Substitute
u = x², du = 2x dx - 2
Rewrite
∫ cos(u) du - 3
Integrate
sin(u) + C - 4
Substitute back
sin(x²) + C
Let u = x² so du = 2x dx; the integral becomes ∫ cos(u) du.
Example 5, Integration by parts
∫ x·eˣ dx
x·eˣ − eˣ + C
- 1
Set up
u = x → du = dx; dv = eˣ dx → v = eˣ - 2
Apply formula
uv − ∫ v du = x·eˣ − ∫ eˣ dx - 3
Integrate the second
= x·eˣ − eˣ + C
∫ u dv = uv − ∫ v du. Choose u = x (becomes simpler when differentiated), dv = eˣ dx.
Step-by-step integral calculator
Integration is the second pillar of calculus. This page covers the main techniques you'll meet in a calculus course: power rule, u-substitution, integration by parts, and definite integrals. Each worked example shows the full reasoning, including the substitution choice or the parts decomposition.
Power rule for integration
For any real n ≠ −1, ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C. The exception n = −1 corresponds to ∫ (1/x) dx = ln|x| + C. The power rule extends to any polynomial term by term, since integrals distribute over addition.
U-substitution explained
When an integrand contains a function and its derivative (up to a constant), substitution simplifies the integral. Let u = (inner function), compute du, replace, integrate in u, then substitute back. Recognizing when substitution applies is a skill that comes with practice, look for an inner function and its derivative as a factor.
Integration by parts
Integration by parts is the analog of the product rule. The formula ∫ u dv = uv − ∫ v du transforms one integral into another that's hopefully easier. Pick u to be something that gets simpler when differentiated (like x or ln x), and dv to be something easy to integrate (like eˣ or trig functions). The mnemonic LIATE (Logs, Inverse trig, Algebraic, Trig, Exponential) helps prioritize u.
Definite integrals and the Fundamental Theorem
The Fundamental Theorem of Calculus says: if F is an antiderivative of f, then ∫ₐᵇ f(x) dx = F(b) − F(a). This turns area computations into algebra, find any antiderivative, plug in the bounds, subtract. The result is a number representing signed area between curve and x-axis.
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