# Seemingly easy integral

I came to this integral which I thaught would be easy to solve, but I'm stuck.

int dx/{x *sqrt{(x ^ 2 )+ k}}
(Sorry Im trying to get it in Latex)

I tried substitution:
$$x ^ 2 + k = t$$
as I remembered from the book (years ago)

But have ended with this:
1/2 times
int dt/{(t-k)sqrt(t)}

Im not sure if this leads to a solution. Any help would be appreciated

Last edited:

## Answers and Replies

Try the substitution x=k1/2tan(y)

Thanks d leet. Sorry I tried in Latex and kept getting most unusual expressions. This is the correct expression.

dextercioby
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Try the substitution x=k^{1/2} sinh t.

Daniel.

that substitution led me to
int dt/sinh t
Any idea how this could be solved?

xAxis said:
that substitution led me to
int dt/sinh t
Any idea how this could be solved?

It shouldn't lead you to that, can you show the work you did using that substitution?

Sure if you don't mind my ascii typing :)
dx = (sqrt)k cosht dt
I = (sqrt)k*int cosht*dt/((sqrt)k*sinht*(sqrt)(k(sinh^2t + 1))) =

=(sqrt)k*int cosht*dt/((sqrt)k*sinht*(sqrt)(k*cosh^2t)) =

==(sqrt)k*int cosht*dt/(k*sinht*cosht)

Then after canceling cosht and puting 1/k out, Im left with dt/sinht

xAxis said:
Sure if you don't mind my ascii typing :)
dx = (sqrt)k cosht dt
I = (sqrt)k*int cosht*dt/((sqrt)k*sinht*(sqrt)(k(sinh^2t + 1))) =

This line is almost impossible to follow, but it doesn't look correct at all. you should have k.5*cosht/(ksinh2 + k).5 inside of the integral.

I got the same under the root of the denominator. When you factor out k, u got (k(sinh^2 + 1))^.5 But that's only the root. There was an x multilplying that root.
I think I reduced it properly, but don't know how to solve dt/sinht

xAxis said:
I got the same under the root of the denominator. When you factor out k, u got (k(sinh^2 + 1))^.5 But that's only the root. There was an x multilplying that root.
I think I reduced it properly, but don't know how to solve dt/sinht

Sorry about that I didn't see the factor of x in the denominator. Well isn't 1/sech(t)= csch(t)? and I think you can look that up in a table of integrals, or I;m sure theres a way to do it, but I'm not sure how do it since I never have before.

dextercioby
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that substitution led me to
int dt/sinh t
Any idea how this could be solved?

Yes, just multiply both the denominator and the numerator by #sinh t# and then use the ch^2 -sh^2 =1 identity.

Daniel.

Thanks dextercioby, your first tip was good as I managed to reduce to the above. As I didn't find it in tables, I used an excelent web integrator
http://integrals.wolfram.com/index.jsp
The solution is ln(tanh(t/2))

dextercioby
Science Advisor
Homework Helper
There's also the substitution

$$x^{2}+1 =p^{2}$$

that leads immediately to a result.

Daniel.

benorin
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Gold Member
Thanks dextercioby, your first tip was good as I managed to reduce to the above. As I didn't find it in tables, I used an excelent web integrator
http://integrals.wolfram.com/index.jsp
The solution is ln(tanh(t/2))

www.quickmath.com is an excellent alternative website for indefinite/definite integration, derivatives, plots, solving equations, matricies, etc.

dextercioby
Science Advisor
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I agree quickmath can perform complicated definite integrals, while Mathematica's web integrator can't.

Daniel.

arildno
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Dearly Missed
Yes, just multiply both the denominator and the numerator by #sinh t# and then use the ch^2 -sh^2 =1 identity.

Daniel.

Alternatively, use a u=tanh(t/2) substitution for the fun of it.