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Latest revision as of 16:09, 26 February 2015
Indefinite Integrals with $ e^x $
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| $ \int e^{ax}dx=\frac{e^{ax}}{a} +C $ |
| $ \int x e^{ax}dx=\frac{e^{ax}}{a}\left(x-\frac{1}{a} \right) +C $ |
| $ \int x^2 e^{ax}dx=\frac{e^{ax}}{a}\left(x^2-\frac{2x}{a}+\frac{2}{a^2}\right) +C $ |
| $ \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 )+C, \qquad n\in {\mathcal N} $ |
| $ \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 +C $ |
| $ \int \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 +C $ |
| $ \int \frac {dx}{p+qe^{ax}}=\frac {x}{p}-\frac {1}{ap} \ln {\left (p+qe^{ax}\right)} +C $ |
| $ \int \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)} +C $ |
| $ \int \frac {dx}{pe^{ax}+qe^{-ax}}= \begin{cases} \frac {1}{a \sqrt{pq}} \arctan {\left ( \sqrt \frac {p}{q}e^{ax} \right)} +C\\ \frac {1}{2a \sqrt{-pq}} \ln {\left( \frac{e^{ax}-\sqrt{-q/p}}{e^{ax}+\sqrt{-q/p}} \right)}+C \\ \end{cases} $ |
| $ \int e^{ax} \sin bx dx = \frac {e^{ax}(a \sin bx-b \cos bx)}{a^2+b^2} +C $ |
| $ \int e^{ax} \cos bx dx = \frac {e^{ax}(a \cos bx-b \sin bx)}{a^2+b^2} +C $ |
| $ \int x e^{ax} \sin bx dx = \frac {x e^{ax}(a \sin bx - b \cos bx)}{a^2+b^2} - \frac {e^{ax} \left \{ (a^2-b^2)\sin bx -2ab \cos bx \right \} }{(a^2+b^2)^2} +C $ |
| $ \int x e^{ax} \cos bx dx = \frac {x e^{ax}(a \cos bx + b \sin bx)}{a^2+b^2} - \frac {e^{ax} \left \{ (a^2-b^2)\cos bx + 2ab \sin bx \right \} }{(a^2+b^2)^2} +C $ |
| $ \int e^{ax} \ln {x} dx = \frac {e^{ax} \ln {x}}{a}-\frac {1}{a} \int \frac {e^{ax}}{x}dx +C $ |
| $ \int e^{ax}\sin^n bx dx = \frac{e^{ax} \sin^{n-1}bx}{a^2+n^2 b^2}(a \sin bx -nb \cos bx) + \frac {n(n-1)b^2}{a^2+n^2 b^2} \int e^{ax} \sin^{n-2} bx dx +C $ |
