Think of an integral as a structured way to ask: “What function has this derivative?”. Once you know a small set of antiderivative patterns (power, exponential, trig), you can solve many integrals quickly—and for harder ones, you choose the right technique (like substitution or integration by parts).
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In most homework and exams, “solve the integral” means: find an antiderivative. If you see an integral sign with no bounds, like ∫ f(x) dx, it is an indefinite integral. Your answer is a family of functions because many different functions have the same derivative.
That + C is not decoration: it represents the fact that derivatives “lose” constants (the derivative of any constant is 0). If the problem gives an initial condition (like F(0) = 2), you can solve for C.
As a reference for standard antiderivative rules and examples, many students like the concise explanations in OpenStax Calculus (antiderivatives and indefinite integration). You don’t need to memorize every trick—just learn to recognize the most common patterns.
A big chunk of integral practice is pattern recognition. Here are the most-used building blocks. (Assume n ≠ −1 in the power rule.)
Two quick reminders:
1) Constants can come out: ∫ 7f(x) dx = 7∫ f(x) dx.
2) “Divide by the new exponent” is the power-rule memory hook.
Problem: Compute ∫ (3x^2 − 4x + 6) dx.
Integrate term-by-term using the power rule:
Check: Differentiate x^3 − 2x^2 + 6x and you get 3x^2 − 4x + 6, so it matches.
Substitution is the go-to move when you see a function and (up to a constant) its derivative sitting next to it. The pattern looks like:
In words: if the integrand has an “inside” expression (like 3x − 1, x^2 + 4, sin(x), etc.), try setting that inside expression equal to u.
Problem: Compute ∫ 2x · (x^2 + 5)^7 dx.
The “inside” is x^2 + 5, and its derivative is 2x, which we already have. Let u = x^2 + 5, then du = 2x dx.
Check idea: Differentiate (x^2+5)^8/8 using the chain rule; you’ll get2x(x^2+5)^7.
Integration by parts is built from the product rule. It’s most useful when you have a product where one factor becomes simpler when you differentiate it (like a polynomial) and the other is easy to integrate (like e^x or sin(x)).
A practical selection tip: choose u so that du is simpler than u. Common choices for u: polynomials (x, x^2, ...), logarithms, inverse trig.
Problem: Compute ∫ x e^x dx.
Choose u = x (so du = dx) and dv = e^x dx(so v = e^x). Then apply the formula.
Check: Differentiate e^x(x − 1). Using product rule, you get xe^x, which matches.
Indefinite integrals represent a family of functions. Always include + C unless the problem is definite (has bounds) or gives an initial condition that determines C.
The power rule does not work for 1/x. Remember: ∫ 1/x dx = ln|x| + C.
If you set u = x^2 + 5, you must replace 2x dx with du. If you can’t rewrite the integrand fully in terms of u and du, the substitution isn’t finished.
Differentiation is the fastest correctness test. If you have time for one extra step, do the derivative check.
For each one, write the technique you chose (basic / substitution / parts) and then do a quick derivative check. That habit is what turns “I hope it’s right” into “I know it’s right.”
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