∫dy/(1+y^2)^1/3

Are you sure that's not:

\int\frac{1}{\sqrt{1+y^{2}}}dy=sinh^{-1}(y)?

I ask because \int\frac{1}{\sqrt[3]{1+x^{2}}}dy can not be integrated using the usual means.

Assuming that Galactus is correct, and what you want to prove is

\int \frac {dy} {\sqrt{1+y^2}} = \sinh ^{-1} y


Here's a general overview of the proof - you can fill in the details:

Start with the equation:
y = \sinh(x)


Take it's derivative, and then use the identity

\cosh^2x -sinh^2x =1

to convert the derivative to a function of \sinh x. Replace the \sinh x terms with y. Rearrange a couple of terms to get dx on one side and the y terms on the other, and integrate. This gives you x is equal to an integral of y. But x = \sinh ^{-1} y; substitute that, and you're done.

If you like, here is a long way to go about it.

\int\frac{1}{\sqrt{1+x^{2}}}dx

Let x=tan(t), \;\ dx=sec^{2}(t)dt

Make the subs and it becomes:

\int \frac{sec^{2}(t)}{\sqrt{\underbrace{1+tan^{2}(t)}_ {\text{sec^2(t)}}}}dt

\int sec(t)dt

This can be evaluated, but it is

ln(sec(t)+tan(t))

Resub t=tan^{-1}(x)

Then, it becomes:

ln(x+\sqrt{1+x^{2}})

This can be shown to be equal to sinh^{-1}(x)

y=sinh^{-1}(x)

y=\frac{e^{y}-e^{-y}}{2}

e^{y}-2x-e^{-y}=0

Multiply by e^{y}

e^{2y}-2xe^{y}-1=0

Using the quadratic formula gives:

e^{y}=x\pm\sqrt{x^{2}+1}

Since e^{y}>0, disregrard the negative solution.

Take logs and we get:

y=ln(x+\sqrt{x^{2}+1})=sinh^{-1}(x)

If you like, here is a long way to go about it.
...

Wow - that is a different way of doing it. But I still like the short way:


y = \sinh(x) \\
\frac {dy} {dx} \ =\ \cosh(x)\ = \ \sqrt{1+\sinh^2x} \ = \ \sqrt{1+y^2} \\
\frac {dy} {\sqrt{1+y^2}} = dx \\
\int \frac {dy} {\sqrt{1+y^2}} = \int dx \ = \ x \ = \sinh^{-1} y

Yes, me too.:)

Why don't you find it helpful? I just thought I would show you another way. It may not be the simplest, but it shows there is always more than one way to tackle a problem. Do you know what I find not helpful? Someone who can't post the correct problem, so we have to interpret. Don't worry, you will not find me helping you anymore.