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Notes

Questions

Find each indefinite integral

$\mathbf{\text{1)}}$ ${\displaystyle \int x+5dx}$

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The answer is ${\displaystyle \frac{x^2}{2}+5x+C}$

$\mathbf{\text{2)}}$ ${\displaystyle \int x^4-2x^2+5xdx}$

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The answer is ${\displaystyle \frac{x^5}{5}-{\displaystyle \frac{2x^3}{3}+{\displaystyle \frac{5x^2}{2}+C}}}$

$\mathbf{\text{3)}}$ ${\displaystyle \int x^{1.5}+5xdx}$

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The answer is ${\displaystyle \frac{2x^{2.5}}{5}+{\displaystyle \frac{5x^2}{2}+C}}$

$\mathbf{\text{4)}}$ ${\displaystyle \int {\displaystyle \frac{1}{x^5}dx}}$

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The answer is ${\displaystyle \frac{-1}{4x^4}+C}$

$\mathbf{\text{5)}}$ ${\displaystyle \int \sqrt[3]{x}+2dx}$

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The answer is ${\displaystyle \frac{3x^{4/3}}{4}+2x+C}$

$\mathbf{\text{6)}}$ ${\displaystyle \int {\displaystyle \frac{x^3+5x^2-4}{\sqrt{x}}dx}}$

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The answer is ${\displaystyle \frac{2x^{3.5}}{7}+2x^{2.5}-8\sqrt{x}+C}$

$\mathbf{\text{7)}}$${\displaystyle \int x^3\sqrt{x}dx}$

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The answer is ${\displaystyle \frac{2x^{4.5}}{9}+C}$

$\mathbf{\text{8)}}$${\displaystyle \int dx}$

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The answer is $x+C$

$\mathbf{\text{9)}}$ ${\displaystyle \int \frac{1}{x}dx}$

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The answer is ${\displaystyle \ln |x|+C}$

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$\text{Since }\frac{d}{dx}\ln |x|=\frac{1}{x},\text{ we have }\int \frac{1}{x}dx=\ln |x|+C.$

$\mathbf{\text{10)}}$ ${\displaystyle \int e^{x}dx}$

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The answer is ${\displaystyle e^x+C}$

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$\text{Since }\frac{d}{dx}{e}^x=e^x,\text{ we have }\int e^{x}dx=e^x+C.$

$\mathbf{\text{11)}}$ ${\displaystyle \int \sin xdx}$

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The answer is ${\displaystyle -\cos x+C}$

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$\text{Since }\frac{d}{dx}(-\cos x)=\sin x,\text{ we have }\int \sin xdx=-\cos x+C.$

$\mathbf{\text{12)}}$ ${\displaystyle \int \cos xdx}$

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The answer is ${\displaystyle \sin x+C}$

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$\text{Since }\frac{d}{dx}\sin x=\cos x,\text{ we have }\int \cos xdx=\sin x+C.$

$\mathbf{\text{13)}}$ ${\displaystyle \int {\sec }^{2}xdx}$

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The answer is ${\displaystyle \tan x+C}$

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$\text{Since }\frac{d}{dx}\tan x={\sec }^{2}x,\text{ we have }\int {\sec }^{2}xdx=\tan x+C.$

$\mathbf{\text{14)}}$ ${\displaystyle \int {\csc }^{2}xdx}$

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The answer is ${\displaystyle -\cot x+C}$

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$\text{Since }\frac{d}{dx}(-\cot x)={\csc }^{2}x,\text{ we have }\int {\csc }^{2}xdx=-\cot x+C.$

$\mathbf{\text{15)}}$ ${\displaystyle \int \sec x\tan xdx}$

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The answer is ${\displaystyle \sec x+C}$

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$\text{Since }\frac{d}{dx}\sec x=\sec x\tan x,\text{ we have }\int \sec x\tan xdx=\sec x+C.$

$\mathbf{\text{16)}}$ ${\displaystyle \int \csc x\cot xdx}$

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The answer is ${\displaystyle -\csc x+C}$

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$\text{Since }\frac{d}{dx}(-\csc x)=\csc x\cot x,\text{ we have }\int \csc x\cot xdx=-\csc x+C.$

$\mathbf{\text{17)}}$ ${\displaystyle \int \frac{1}{1+x^2}dx}$

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The answer is ${\displaystyle {\tan }^{-1}x+C}$

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$\text{Since }\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2},\text{ we have:}$

${\displaystyle \int \frac{1}{1+x^2}dx={\tan }^{-1}x+C}$

$\mathbf{\text{18)}}$ ${\displaystyle \int \frac{1}{\sqrt{1-x^2}}dx}$

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The answer is ${\displaystyle {\sin }^{-1}x+C}$

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$\text{Since }\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}},\text{ we have:}$

${\displaystyle \int \frac{1}{\sqrt{1-x^2}}dx={\sin }^{-1}x+C}$

See Related Pages$$

$\bullet \text{ Indefinite Integral Calculator }$
$\text{(Symbolab.com)}$

$\bullet \text{ Calculus Homepage}$
$\text{All the Best Topics…}$

$\bullet \text{ Trapezoidal Rule}$
${\displaystyle \frac{b-a}{2n}[f(a)+2f(x_1)+2f(x_2)+\dots +2f{x}_{n-1}+f(b)]\dots }$

$\bullet \text{ Properties of Integrals}$
${\displaystyle {\int }_a^{b}cf(x)dx=c{\displaystyle {\int }_a^{b}f(x)dx\dots }}$

$\bullet \text{ Indefinite Integrals- Power Rule}$
${\displaystyle \int x^{n}dx={\displaystyle \frac{x^{n+1}}{n+1}+C\dots }}$

$\bullet \text{ Indefinite Integrals- Trig Functions}$
${\displaystyle \int \cos xdx=\sin x+C\dots }$

$\bullet \text{ Definite Integrals}$
${\displaystyle {\int }_5^7{x}^{3}dx\dots }$

$\bullet \text{ Integration by Substitution}$
${\displaystyle \int (x^2+3{)}^3(2x)dx\dots }$

$\bullet \text{ Area of Region Between Two Curves}$
$A={\displaystyle {\int }_a^b[f(x)-g(x)]dx\dots }$

$\bullet \text{ Arc Length}$
${\displaystyle {\int }_a^b\sqrt{1+{[f^{\prime }(x)]}^2}dx\dots }$

$\bullet \text{ Average Function Value}$
${\displaystyle \frac{1}{b-a}{\int }_a^{b}f(x)dx}$

$\bullet \text{ Volume by Cross Sections}$
$$

$\bullet \text{ Disk Method}$
$V={\displaystyle {\int }_a^b{[f(x)]}^{2}dx\dots }$

$\bullet \text{ Cylindrical Shells}$
$V=2\pi {\displaystyle {\int }_a^{b}yf(y)dy\dots }$

In Summary

Indefinite integrals, also known as antiderivatives, are a key concept in calculus. Eventually we will work our way up to finding definite integrals which will enable us to do many more things with functions, for example, finding the area under a curve.

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