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! style="background-color: rgb(228, 188, 126); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Table of Infinite Integrals Continues
 
! style="background-color: rgb(228, 188, 126); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Table of Infinite Integrals Continues
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! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | 27 Integrals Component <math> e^{ax} </math>
 
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|<math> \int e^{ax}dx=\frac{e^{ax}}{a} </math>
 
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|<math> \int x e^{ax}dx=\frac{e^{ax}}{a}\left(x-\frac{1}{a} \right) </math>
 
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|<math> \int x^2 e^{ax}dx=\frac{e^{ax}}{a}\left(x^2-\frac{2x}{a}+\frac{2}{a^2}\right) </math>
 
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|<math> \int x^n e^{ax}dx=\frac{x^n e^{ax}}{a}-\frac{n}{a} \int x^{n-1} e^{ax}dx = \frac {e^{ax}}{a} \left( x^n- \frac{nx^{n-1}}{a}+\frac{n(n-1)x^{n-2}}{a^2}- \cdot \cdot \cdot \frac{(-1)^n n!}{a^n} \right ) \qquad \text{if n is a poaitive integer} </math>
 
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<math> \int \frac {e^{ax}}{x}dx=\ln {x} + \frac {ax}{1 \cdot 1!} + \frac {(ax)^2}{2 \cdot 2!} + \frac {(ax)^3}{3 \cdot 3!} + \cdot \cdot \cdot </math>
 
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<math> \frac {e^{ax}}{x^n}dx = \frac {-e^{ax}}{(n-1)x^{n-1}} + \frac {a}{n-1} \int \frac {e^{ax}}{x^{n-1}}dx </math>
 
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<math> \frac {dx}{p+qe^{ax}}=\frac {x}{p}-\frac {1}{ap} \ln {\left (p+qe^{ax}\right)} </math>
 
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<math> \frac {dx} {\left ( p+qe^{ax} \right) ^2}=\frac {x}{p^2}+\frac {1}{ap(p+qe^{ax})} -\frac{1}{ap^2}\ln {\left (p+qe^{ax}\right)} </math>
 
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Latest revision as of 12:57, 24 November 2010


Table of Infinite Integrals Continues
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Ruth Enoch, PhD Mathematics