seemingly easy integral

[xAxis]

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

[d_leet]

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

[xAxis]

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

[dextercioby]

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

Daniel.

[xAxis]

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

[d_leet]

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?

[xAxis]

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

[d_leet]

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.

[xAxis]

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

[d_leet]

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]

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.

[xAxis]

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]

There's also the substitution

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

that leads immediately to a result.

Daniel.

[benorin]

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]

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

Daniel.

[arildno]

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.