Is There a Simple Solution for This Integral?

[MathewsMD]

for the problem:

\int {[(y-1)^3 + C]^{-1/2}}dy

Is there a simple solution that can yield an answer? C is a constant. Integration by parts doesn't seem to look helpful (at least to me). Trigonometric substitution looks like one method that would work, though it would involve quite a bit of substitution and it appears it will be messy. Are there any other methods that are simpler and would work well?

 

[MathewsMD]

Here's my solution. I think it suffices in saying that the answer isn't exactly as simplified as it should be. The top is kind of cut out, but it reads

\frac {1}{c^{1/2}}(y - 1)^{3/2} = u

Any help with regards to finding a simpler solution would be greatly appreciated!

 

[Mark44]

And the bottom is cut off as well.

You can always verify that an antiderivative is correct by differentiating it. If you wind up with the integrand, then all is well.

 

[MathewsMD]

And the bottom is cut off as well.

You can always verify that an antiderivative is correct by differentiating it. If you wind up with the integrand, then all is well.

The bottom is merely substitution. Yes, verifying with the antiderivative to see if it a solution is one possibility, but I was wondering for a question like this, if there was another method besides trigonometric substitution that would help simplify the integration. Any help is greatly appreciated!

 

[Mark44]

This problem doesn't lend itself easily to a trig substitution, IMO. That type of substitution is most useful when you have a sum or difference of squares, raised to some power. Having a cube in there as you do makes a trig sub less promising. Maybe you can make it work, treating it as a sum of squares -- [(y + 1)^3/2]² + (sqrt(C))²

And maybe that's what you did, but without being able to see it all, I didn't put any time into checking.

 

[MathewsMD]

This problem doesn't lend itself easily to a trig substitution, IMO. That type of substitution is most useful when you have a sum or difference of squares, raised to some power. Having a cube in there as you do makes a trig sub less promising. Maybe you can make it work, treating it as a sum of squares -- [(y + 1)^3/2]² + (sqrt(C))²

And maybe that's what you did, but without being able to see it all, I didn't put any time into checking.

Sorry, I thought I posted this earlier...there must have been some kind of mess-up on my part. This is what I was referring to as revised. The solution doesn't seem to be easily solvable for its coefficients, and trig substation, although messy, seems like the only elementary way to go about the problem, no?

 

[Mark44]

The first attached image, soln.jpg, is blurry and unreadable in part. The second image, soln2.jpg, is cut off again.

 

[dextercioby]

I'm pretty sure this is doable in terms of elliptic integrals.

http://www.wolframalpha.com/input/?i=Integrate+1/(x^3+1)^(1/2)