Eksponensial integral
[HMMT 2008] Jika 
Solusi
Pecahan tadi dapat diubah menjadi
Integralnya adalah
![\displaystyle T=\int_{0}^{\ln 2} \frac{u'(x)dx}{u(x)} = \int_{u(0)}^{u(\ln 2} \frac{du}{u(x)} = \left[\ln |u(x)| \right]_{x=0}^{\ln 2}](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+T%3D%5Cint_%7B0%7D%5E%7B%5Cln+2%7D+%5Cfrac%7Bu%27%28x%29dx%7D%7Bu%28x%29%7D+%3D+%5Cint_%7Bu%280%29%7D%5E%7Bu%28%5Cln+2%7D+%5Cfrac%7Bdu%7D%7Bu%28x%29%7D+%3D+%5Cleft%5B%5Cln+%7Cu%28x%29%7C+%5Cright%5D_%7Bx%3D0%7D%5E%7B%5Cln+2%7D&bg=ffffff&fg=000000&s=0 "\displaystyle T=\int_{0}^{\ln 2} \frac{u'(x)dx}{u(x)} = \int_{u(0)}^{u(\ln 2} \frac{du}{u(x)} = \left[\ln |u(x)| \right]_{x=0}^{\ln 2}")
Substitusikan nilai 
![\displaystyle T=\left[ \ln |e^{2x}+e^x + e^{-x}-1| \right]_{x=0}^{\ln 2} = \ln \frac{11}{2} - \ln 2 = \ln \frac{11}{4}](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+T%3D%5Cleft%5B+%5Cln+%7Ce%5E%7B2x%7D%2Be%5Ex+%2B+e%5E%7B-x%7D-1%7C+%5Cright%5D_%7Bx%3D0%7D%5E%7B%5Cln+2%7D+%3D+%5Cln+%5Cfrac%7B11%7D%7B2%7D+-+%5Cln+2+%3D+%5Cln+%5Cfrac%7B11%7D%7B4%7D&bg=ffffff&fg=000000&s=0 "\displaystyle T=\left[ \ln |e^{2x}+e^x + e^{-x}-1| \right]_{x=0}^{\ln 2} = \ln \frac{11}{2} - \ln 2 = \ln \frac{11}{4}")
Maka 
3 Mei 2008
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