Supplement: Common Antiderivatives

Formula reference. Differentiate any proposed antiderivative to check it. Each formula holds on an interval where its integrand is defined.

Common antiderivatives

$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\ (n\ne-1),\qquad \int\frac1x\,dx=\ln|x|+C,\qquad \int e^x\,dx=e^x+C.$

$\int a^x\,dx=\frac{a^x}{\ln a}+C\ (a>0,\ a\ne1),\qquad \int\cos x\,dx=\sin x+C,\qquad \int\sin x\,dx=-\cos x+C.$

$\int\sec^2x\,dx=\tan x+C,\qquad \int\csc^2x\,dx=-\cot x+C,\qquad \int\sec x\tan x\,dx=\sec x+C.$

$\int\csc x\cot x\,dx=-\csc x+C,\qquad \int\frac1{1+x^2}\,dx=\arctan x+C,\qquad \int\frac1{\sqrt{1-x^2}}\,dx=\arcsin x+C.$

An antiderivative is best remembered as a function whose derivative returns the integrand. That definition turns every reference entry into something you can check. If you claim that an expression is an antiderivative of $f(x)$, differentiate your claim. The check is quick, local, and more reliable than recall alone.

Imagine reconstructing a train’s position from a recorded velocity. Many position functions can have the same velocity because adding a constant vertical shift does not change a derivative. The initial position selects one member of that family. This story explains why indefinite integrals require $+C$ and why a definite integral does not.

Recover powers, exponentials, and logarithms

For $n\ne-1$, $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$. Differentiate $\frac{x^{n+1}}{n+1}$ to see the coefficient and exponent cancel correctly. The exception is $\int\frac1x\,dx=\ln|x|+C$ on intervals that avoid zero. The absolute value is needed because both positive and negative nonzero inputs have derivative $\frac1x$ after the logarithmic form is interpreted on its interval.

Exponential families reverse cleanly: $\int e^x\,dx=e^x+C$, while $\int a^x\,dx=\frac{a^x}{\ln a}+C$ for $a>0$ and $a\ne1$. The denominator appears because $(a^x)'=(\ln a)a^x$. A missing $\ln a$ is a derivative-check failure, not a minor formatting error.

Recover trigonometric functions with signs

Because $(\sin x)'=\cos x$, $\int\cos x\,dx=\sin x+C$. Because $(\cos x)'=-\sin x$, $\int\sin x\,dx=-\cos x+C$. The minus sign is a consequence of differentiation, and it is worth checking rather than memorizing as an exception.

For secant-squared and cosecant-squared, $\int\sec^2x\,dx=\tan x+C$ and $\int\csc^2x\,dx=-\cot x+C$ on intervals where the expressions are defined. These statements inherit the trig functions’ discontinuities. Choose an interval before treating one antiderivative family as a global formula.

Connect the reference to a model

If a train has velocity $v(t)=3t^2-4\,\frac{\mathrm{m}}{\mathrm{s}}$, then an antiderivative is $s(t)=t^3-4t+C$ meters. The units are consistent because integrating velocity with respect to seconds yields meters. Given $s(0)=12\,\mathrm{m}$, the constant is $12$, producing a particular position model.

Do not confuse that antiderivative with distance traveled. The difference $s(b)-s(a)$ gives displacement, which may cancel when velocity changes sign. Total distance requires integrating $|v(t)|$ or splitting at sign changes. A reference table provides forms; the model determines what the form means.

Narrative challenge: reconstruct the train position

A train has velocity $v(t)=6t-8\,\frac{\mathrm{m}}{\mathrm{s}}$ and position $s(0)=15\,\mathrm{m}$. Before opening the solution, find a position function and explain why the constant cannot be omitted. Then identify the time at which velocity is zero and state why that time may matter for distance.

Show a solution path

An antiderivative is $s(t)=3t^2-8t+C$. The initial condition gives $C=15$, so $s(t)=3t^2-8t+15$ meters. Velocity is zero when $6t-8=0$, or $t=\frac43\,\mathrm{s}$. A sign change there would make it a necessary split point when computing total distance.