razonamiento matematico: octubre 2009

$\int%20\frac{du}{\sqrt{k^2{}+u^2{}}}=L_n{}(u@plus;\sqrt{k^2{}@plus;u^2{}})$

$\int%20\frac{du}{k^2{}-u^2{}}=\frac{1}{2k}L_n{\frac{k+u}{k-u}}$

$\inline%20\int%20ku(x)dx=k\int%20u(x)dx$

$\int\frac{du}{u^2{}-k^2{}}=\frac{1}{2k}\cdot%20L_n{}\frac{u-k}{u+k}$

$\inline%20\int%20u^n{du}=\frac{u^n^+^1}{n@plus;1}{}{}{}$

$\int%20\sin%20^2{u}du=(\frac{1}{2})u-(\frac{1}{4})\sin%20(2u)$

$\inline%20\int%20f(kx)dx=\frac{1}{k}\cdot%20\int%20f(u)du$

$\int\csc%20udu=L_n{}\tan%20(\frac{u}{2})$

$\inline%20\int%20\frac{du}{u}=L_n{}\left%20|%20u%20\right%20|$

$\inline%20\int%20\sec%20udu=L_u{}(\sec%20u+\tan%20u)=L_n{}\tan%20(\frac{u}{2})$

$\inline%20\int%20e^u{du}=e^u{}$

$\inline%20\int%20\frac{\cos%20u}{\sin%20^2{u}}\cdot%20du=-\csc%20u$

$\inline%20\int%20k^u{du}=\frac{k^u{}}{L_nk{}};k%3E0;k\neq%201$

$\inline%20\int%20\frac{\sin%20u}{\cos%20^2{u}}\cdot%20du=\sec%20u$

$\inline%20\int\sqrt{u}du=\frac{u\tfrac3{2}{}}{\frac{3}{2}}=\frac{2\cdot%20u\tfrac3{2}{}}{3}$

$\inline%20\int%20\cot%20u%20du=L_u{\sin%20u}$

$\inline%20\int%20\tan%20u%20du=L_n{}\sec%20u=-L_n{}\cos%20u$

$\inline%20\int%20\sin%20udu=-\cos%20u$

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jueves, 29 de octubre de 2009

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