Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (anonymous):

Use u-substitution: The integral dx/1+e^x

OpenStudy (zzr0ck3r):

\[\int\frac{1}{1+e^x}dx\]?

OpenStudy (anonymous):

yes, I tried using the entire denominator as u but that doesn't seem to work

OpenStudy (anonymous):

unless I did it wrong

OpenStudy (zzr0ck3r):

\[\int\frac{1}{1+e^x}dx=\int\frac{1+e^x-e^x}{1+e^x}dx=\int \frac{1+e^x}{1+e^x}dx-\int\frac{e^x}{1+e^x}=\\\int1dx-\int\frac{e^x}{1+e^x}dx=x-\int\frac{e^x}{1+e^x}dx\]now that we have this, use u = 1+e^x, du = e^x dx so \[x-\int\frac{e^x}{1+e^x}dx=x-\int\frac{1}{u}du=x-\ln(|1+e^x|)+c\]

OpenStudy (zzr0ck3r):

bbiab

OpenStudy (anonymous):

ok, thank you!

OpenStudy (anonymous):

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!

OpenStudy (zzr0ck3r):

adding zero trick is good stuff:)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!