integrating with an e

[dolor]

Does anyone know how to integrate the sqrt(1 + e^x) ? Maybe I just don't see it, but doing it by parts or trig substitution isn't working for me. Any tips? Thanks in advance.

[krtxmrtz]

Let I represent the integral to be solved. And let In{} represent the integral sign.

Now, use this substitution:

1 + e^-x = p²

whereby,

-e^-xdx = 2pdp

and

dx = 2pdp / -e^-x = -2pdp / (p² - 1)

Therefore:

I = -2*In{p²dp / (p² - 1)}

By simple algebra this can be decomposed into:

I = -In{p / (p + 1)} - In{p / (p - 1)}

and these two are easy enough (for example, substitute further p + 1 = v in the first term and p - 1 = w in the second). The rest should be straightforward.

[krtxmrtz]

By the way, the correct solution is:

I = Ln{[Sqrt(1 + e^-x) + 1] / [Sqrt(1 + e^-x) - 1]} - 2*Sqrt(1 + e^-x)

where Ln means natural logarithm (base e)

I hope you don't get confused by the notation.

[dolor]

thanks! i never thought of that substitution.