- #1
Zach Knight
- 11
- 0
Homework Statement
[tex]\int{\sqrt{1+e^x}dx}[/tex]
Homework Equations
[tex]\int{uv'}=uv-\int{u'v}[/tex]
The Attempt at a Solution
I rewrote the integrand as
[tex]\sqrt{1+(e^{x/2})^2}[/tex]
and used the trigonometric substituition [tex]e^{x/2}=tan(\theta)[/tex], which simplified the radical to
[tex]\int{\sqrt{1+e^x}dx}=2\int{csc(\theta)sec^2(\theta)d\theta}[/tex]
From there I used integration by parts, with [tex]u=csc(\theta)[/tex], and [tex]v'=sec^2(\theta)[/tex], which gave me
[tex]2\int{csc(\theta)sec^2(\theta)d\theta}=2(sec(\theta)+\int{sin^2(\theta)d\theta})[/tex]
Solving that, and undoing all of my substitutions, I found that
[tex]\int{\sqrt{1+e^x}dx}=2sec(arctan(e^{x/2}))+arctan(e^{x/2})-\frac{1}{2}sin(arctan(e^{x/2}))[/tex]
WolframAlpha, on the other hand, gives
[tex]\int{\sqrt{1+e^x}dx}=2\sqrt{1+e^x}-2arctanh^{-1}(\sqrt{1+e^x})+A[/tex],
and I have no idea how. Even after rewriting the trig-inverse trig pairs as algebraic expressions, I still don't get close to what Wolfram gave. Can somebody show me where I went wrong?