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Calcul de integrala 6
Posted: Sun Feb 10, 2008 12:19 am
by ins-
Ce se calculeze: \( \int \frac {1-x}{e^{x}+x^{2}e^{-x}}\ dx \).
Posted: Sun Feb 10, 2008 7:11 pm
by Doru Popovici
\( I=\int (1-x)dx/(e^x+x^2e^{-x}) \)
\( I=\int (1-x)e^{-x}dx/(1+(x/e^x)^2) \)
Vom face o schimbare de variabila:
\( t=x/e^x\Rightarrow dt=(1-x)e^{-x}dx \)
=>\( I=\int dt/(1+t^2)=tan^{-1} t +C \)
=>\( I=tan^{-1}(xe^{-x}) + C \)