Finding the Indefinite Integral of a Radical Expression

[Zach Knight]

Homework Statement

\int{\sqrt{1+e^x}dx}

Homework Equations

\int{uv'}=uv-\int{u'v}

The Attempt at a Solution

I rewrote the integrand as
\sqrt{1+(e^{x/2})^2}
and used the trigonometric substituition e^{x/2}=tan(\theta), which simplified the radical to
\int{\sqrt{1+e^x}dx}=2\int{csc(\theta)sec^2(\theta)d\theta}
From there I used integration by parts, with u=csc(\theta), and v'=sec^2(\theta), which gave me
2\int{csc(\theta)sec^2(\theta)d\theta}=2(sec(\theta)+\int{sin^2(\theta)d\theta})
Solving that, and undoing all of my substitutions, I found that
\int{\sqrt{1+e^x}dx}=2sec(arctan(e^{x/2}))+arctan(e^{x/2})-\frac{1}{2}sin(arctan(e^{x/2}))
WolframAlpha, on the other hand, gives
\int{\sqrt{1+e^x}dx}=2\sqrt{1+e^x}-2arctanh^{-1}(\sqrt{1+e^x})+A,
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?

 

[Mentallic]

You can simplify such expressions as \sin\left(\sin^{-1}(x)\right)=x and \sin\left(\cos^{-1}(x)\right)=\sqrt{1-x^2}
Similarly, you can simplify your expressions.

 

[jhae2.718]

They are equivalent.

Wolfram|Alpha's answer should be in terms of arctanh, not arctanh^-1.

 

[Mentallic]

Yep. As a rule of thumb, if you have a different answer to the answer in the back of the book or wolfram alpha, you should always check to see if they truly are different.

It's as though you're telling me the answer is 2² but I'm telling you the answer is 4, and then you start worrying that your answer is wrong

 

[Zach Knight]

They are equivalent.

Wolfram|Alpha's answer should be in terms of arctanh, not arctanh^-1.

Sorry, I combined arctanh and tanh^-1
Anyway, how are the two equivalent? I'm not very familiar with hyperbolic trigonometry.

 

[jhae2.718]

Actually, I think I may be mistaken. (It's late here!)

I'll do the integral and post what I get.

 

[ideasrule]

From there I used integration by parts, with u=csc(\theta), and v'=sec^2(\theta), which gave me
2\int{csc(\theta)sec^2(\theta)d\theta}=2(sec(\theta)+\int{sin^2(\theta)d\theta})

This step is not correct. The integral of sec^2(theta) is tan(theta) and the derivative of csc(theta) is -csc(theta)*cot(theta), so the product of the two should be -csc(theta).

 

[jhae2.718]

I get:
\int\sqrt{1+e^x}dx = 2\sqrt{1+e^x} + \ln\left|\frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|+C.

 

[Zach Knight]

This step is not correct. The integral of sec^2(theta) is tan(theta) and the derivative of csc(theta) is -csc(theta)*cot(theta), so the product of the two should be -csc(theta).

Ah, that's it! Somehow, I got the derivative of csc(theta) as -sin(theta)cos(theta). I guess it's time for me to go to bed