Question

Hi,
Could someone walk me through this question:

Evaluate the improper integral sinh2x/1+cosh^4x dx between 0 and infinity.
So far I have changed sinh2x to 2sinhxcoshx and I think I will use u substitution and make u=coshx

Also du=sinhx, so I have u.du/1+u^4...does that mean I integrate to get (u^2/2) / x+u^5/5 ?

And then I get to 5u^2/x+u^5...

Then for x i have cosh-1u?

So overall it is 5u^2/cosh-1u+u^5