Question

Evaluate the integral:

Answer 1

@Kainui

Answer 2

Integral of @Kainui = infinity.

Answer 3

http://integrals.wolfram.com/index.jsp?expr=&random=false

Answer 4

\[\int\limits_{0}^{1}\frac{ \sqrt{1+(e^x)} }{ e^x }dx\]

Answer 5

the numerator e^x should be negative so 1+ e^-x

Answer 6

Wait, so the integral is really:

\[\int\limits_{0}^{1} \frac{ \sqrt{1+e^{-x}} }{ e^{-x} }\]

Answer 7

Just the numerator is ^-x it wouldn't work for me when I tried haha

Answer 8

This was my first question I think haha..just remembered (2 am, ofc this would happen)

Answer 9

Ahh, well then it looks like this:

\[\int\limits_{0}^{1} e^{-x} \sqrt{1+e^{-x}} \]

right?

Answer 10

yup

Answer 11

Ok, and I guess I forgot the dx at the end, but basically it should be a breeze if you choose 1+e^(-x)=u.

Answer 12

:P

Answer 13

Solve it!