Notes
Before you start: differentiate powers, simplify indices and substitute into an expression.
An antiderivative F satisfies F′=f on the interval being considered. The indefinite integral describes the family F+C, because differentiating a constant gives zero. On a connected interval any two antiderivatives differ by a constant; separate intervals of a disconnected domain need not share one constant.
\[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\qquad(n\ne-1).\]
Increase the power and divide by the new power. Integrate sums term by term and keep constant multipliers. Negative powers require x≠0; fractional-power examples here use positive bases. The n=−1 case cannot use this formula: it would divide by zero and requires a logarithmic antiderivative.
For a nonzero constant a and n≠−1, reverse the linear-inner-function chain rule:
\[\int(ax+b)^n\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}+C.\]
Check by differentiating. If a=0 and the integrand is defined, integrate the resulting constant instead of dividing by a. For a continuous integrand on [u,v] with antiderivative F throughout the interval,
\[\int_u^v f(x)\,dx=F(v)-F(u).\]
Upper limit first, lower limit second. A constant cancels, so a definite integral is a number without an arbitrary +C. Reversing limits changes its sign. Check the entire interval: endpoint subtraction alone does not justify crossing a singularity.
Recall later
\(\int 3x^2\,dx=x^3+C\), while \(\int_0^2 3x^2\,dx=8\). The two requests have different types of answer.
Common mistakes
- Using the differentiation power rule instead of increasing the exponent.
- Omitting an indefinite constant or adding an arbitrary constant to a definite answer.
- Forgetting the inner-function divisor, using the n=−1 formula or crossing a zero denominator.