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Calculus1
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Evaluate the follow integral and simplify the result. ∫1/((2e^x) -3)
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\[\int \frac{1}{2e^x-3}dx\]??
Yes, can you help me?
let \(u=e^x\) => du/dx=e^{x}
\[\int \frac{1}{2u-3}*e^{x}du =>\int \frac{1}{u(2u-3)}du\]
Wait, I thought that if du/dx=e^{x}, then dx=du/e^x.
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\[\frac{du}{dx}=e^x\] \[du=e^{x}dx\] \[e^{x}dx=du\] \[dx=\frac{du}{e^x}\]
yeah, i made a typo..but the last part is still right.
So is it \[\int\limits_{}^{}\frac{ 1 }{ 2e^x-3 } \frac{ du }{ e^x }\]
u=e^x
then proceed with partial fractions.
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