Solve the integral in terms of inverse hyperbolic functions

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rir0302

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∫1/(1-17x^2) dx with limits [4,5]

I know ∫1/(1-x^2) dx is coth^-1(x) but what do I do with the 17?
 
Try a substitution. Where you have 17x^2, you wish you have just u^2. What should u be?

Be sure to show your work, so we can tell if you are handing the details correctly.
 
Ohh, so

u=17x
du=17dx
1/17∫1/(1-u^2) dx
1/17(coth^-1(u))
1/17(coth^-1(17x))
On [4,5]:
1/17(coth^-1(17*5) - coth^-1(17*4))
1/17(coth^-1(85) - coth^-1(68))

Is this right?
 
rir0302 said:
Ohh, so

u=17x
du=17dx
1/17∫1/(1-u^2) dx
1/17(coth^-1(u))
1/17(coth^-1(17x))
On [4,5]:
1/17(coth^-1(17*5) - coth^-1(17*4))
1/17(coth^-1(85) - coth^-1(68))

Is this right?
Click to expand...
No.

Your u^2 is not 17x^2, but (17x)^2 = 289x^2. Pay attention!

Also, check whether you copied the problem correctly; that 17 does make the problem rather ugly, so maybe it wasn't intended to be where you put it.
 
Yes, the problem is correct. I'm not sure where the u^2 applies though, sorry - since it integrates into (coth^-1(u))?
 
rir0302 said:
u=17x
du=17dx
1/17∫1/(1-u^2) dx
1/17(coth^-1(u))
1/17(coth^-1(17x))
On [4,5]:
1/17(coth^-1(17*5) - coth^-1(17*4))
1/17(coth^-1(85) - coth^-1(68))

Is this right?
Click to expand...
The problem was ∫1/(1-17x^2) dx . You are claiming that 17x^2 = u^2. But those are not equal, with your substitution! Rather, u^2 will be (17x)^2 = 289x^2, which is not what is in the integrand!

If you want 17x^2 to equal u^2, what should u actually be defined as? It is not 17x.
 
... or convert back to x's and use the original limits of integration

Try both ways to decide which is easier for you!
 
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