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Antiderivatives and Indefinite Integrals

A function FF is an antiderivative of a function ff if F′(x)=f(x)F'\left(x\right)=f\left(x\right)

The general antiderivative or indefinite integral of a function f(x)f\left(x\right) is given by
∫f(x)dx=F(x)+C\displaystyle\int_{ }^{ }f\left(x\right)dx=F\left(x\right)+C
  • CC is any constant
  • The pair of symbols ∫\displaystyle\int_{ }^{ } and dxdx come and go together -- we remove them by performing the action of finding the antiderivative
*The indefinite integral of a function is a function


Properties of an Indefinite Integral
  • ∫c f(x)dx=c∫f(x)dx\displaystyle\int_{ }^{ }c\ f\left(x\right)dx=c\int_{ }^{ }f\left(x\right)dx
  • ∫[f(x)±g(x)]dx=∫f(x)dx ± ∫g(x)dx\displaystyle \int_{ }^{ }\left[f\left(x\right)\pm g\left(x\right)\right]dx=\int_{ }^{ }f\left(x\right)dx\ \pm\ \int_{ }^{ }g\left(x\right)dx
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Common Antiderivatives/integrals

Power Rule∫1 dx=x+C∫xn dx=xn+1n+1+CExp Rule∫ex dx=ex+C∫ax dx=axln⁡a+CLog Rule∫1x=ln⁡∣x∣+CTrig Rule∫sin⁡x dx=−cos⁡x+C∫cos⁡x dx=sin⁡x+C∫sec⁡2x dx=tan⁡x+C∫csc⁡2x dx=−cot⁡x+C∫csc⁡xcot⁡x dx=−csc⁡x+C∫sec⁡xtan⁡x dx=sec⁡x+C∫1a2+x2 dx=1aarctan⁡(xa)+C∫1a2−x2 dx=arcsin⁡(xa)+C\begin{array}{|lll|} \hline \textcolor{green}{\text{Power Rule}}\\ \displaystyle\int 1\ dx=x+C& \displaystyle\int x^n\ dx=\frac{x^{n+1}}{n+1}+C\\ \hline \textcolor{green}{\text{Exp Rule}}\\ \displaystyle\int e^x\ dx=e^x+C& \displaystyle\int a^x\ dx=\frac{a^x}{\ln a}+C\\ \hline \textcolor{green}{\text{Log Rule}}\\ \displaystyle\int \frac{1}{x}=\ln|x|+C&&\\ \hline \textcolor{green}{\text{Trig Rule}}\\ \displaystyle\int \sin x\ dx=-\cos x+C& \displaystyle\int \cos x\ dx=\sin x+C\\ \displaystyle\int \sec^2 x\ dx=\tan x +C& \displaystyle\int \csc^2 x\ dx=-\cot x +C\\ \displaystyle\int \csc x \cot x\ dx=-\csc x+C& \displaystyle\int \sec x \tan x\ dx=\sec x+C \\ \displaystyle\int\frac{1}{a^2+x^2}\ dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C& \displaystyle\int\frac{1}{\sqrt{a^2-x^2}}\ dx=\arcsin\left(\frac{x}{a}\right)+C&\\ \hline \end{array}

Example
Evaluate ∫[2x−34−x2+2sin⁡x]dx\displaystyle \int_{ }^{ }\left[2\sqrt{x}-\frac{3}{\sqrt{4-x^2}}+2\sin x\right]dx
Rewriting:
=2∫x12dx−3∫14−x2dx+2∫sin⁡xdx\displaystyle =2\int_{ }^{ }x^{\frac{1}{2}}dx-3\int_{ }^{ }\frac{1}{\sqrt{4-x^2}}dx+2\int_{ }^{ }\sin xdx
=2x3232−3arcsin⁡(x2)+2(−cos⁡x)+C\displaystyle =2\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-3\arcsin\left(\frac{x}{2}\right)+2\left(-\cos x\right)+C
=43x32−3arcsin⁡(x2)−2cos⁡x+c\displaystyle =\frac{4}{3}x^{\frac{3}{2}}-3\arcsin\left(\frac{x}{2}\right)-2\cos x+c
Evaluate the integral ∫1+1x+2x2+3x3dx\displaystyle\int_{ }^{ }1+\frac{1}{x}+\frac{2}{x^2}+\frac{3}{x^3}dx.
Evaluate the integral∫[tan⁡2x+1+1sec⁡2x+sin⁡2x]dx\displaystyle \int_{ }^{ }\left[\tan^2x+1+\frac{1}{\sec^2x}+\sin^2x\right]dx.
Hint: First simplify using trig identities
Find f(x)f\left(x\right), given that f′(x)=x2−2+sin⁡xf'\left(x\right)=x^2-2+\sin x and f(0)=3f\left(0\right)=3.

Extra Practice