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Use u-substitution: The integral dx/1+e^x
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\[\int\frac{1}{1+e^x}dx\]?
yes, I tried using the entire denominator as u but that doesn't seem to work
unless I did it wrong
\[\int\frac{1}{1+e^x}dx=\int\frac{1+e^x-e^x}{1+e^x}dx=\int \frac{1+e^x}{1+e^x}dx-\int\frac{e^x}{1+e^x}=\\\int1dx-\int\frac{e^x}{1+e^x}dx=x-\int\frac{e^x}{1+e^x}dx\]now that we have this, use u = 1+e^x, du = e^x dx so \[x-\int\frac{e^x}{1+e^x}dx=x-\int\frac{1}{u}du=x-\ln(|1+e^x|)+c\]
bbiab
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ok, thank you!
thank you again, I remember learning about your "trick step" in my mathematical proofs class last semester (adding e^x - e^x in the numerator) but I didn't even think to use it in my calc 2 class. thank you so much!
adding zero trick is good stuff:)
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