challenging integral with square root of a polynomial which includes a negative power of x

[Evari5te]

The original question is "Find the length of the curve 8(y + ln x) = x^2 between x = 1 and x = e.

I rearrange to give y = x^2/8 - ln x.

I understand the approach would be to use S = integral ( 1 + (dy/dx)^2 )^(1/2) dx

I arrive at dy/dx = x/4 - 1/x

(dy/dx)^2 = (x^2)/16 - (1/2) + (1/(x^2))

So adding the 1 the polynomial inside the square root is (x^2)/16 + (1/2) + (1/(x^2)).

The answer given is (e^2 + 7) / 8, but I am stuck as to how to resolve the integration of the square rooted polynomial which includes the x^-2 term.

 

[Deleted member 4993]

The original question is "Find the length of the curve 8(y + ln x) = x^2 between x = 1 and x = e.

I rearrange to give y = x^2/8 - ln x.

I understand the approach would be to use S = integral ( 1 + (dy/dx)^2 )^(1/2) dx

I arrive at dy/dx = x/4 - 1/x

(dy/dx)^2 = (x^2)/16 - (1/2) + (1/(x^2))

So adding the 1 the polynomial inside the square root is (x^2)/16 + (1/2) + (1/(x^2)).

The answer given is (e^2 + 7) / 8, but I am stuck as to how to resolve the integration of the square rooted polynomial which includes the x^-2 term.

I would try substitution first. I have not completed this but I would start with:

x = (√8) * tan(Φ)

and see what do I get ........

 

[limiTS]

Hint:
[math]\frac{x^2}{16}+\frac{1}{2}+\frac{1}{x^2}=(a+b)^2[/math]
It's a perfect square which will take care of the square root.

 

[Steven G]

Hint:
[math]\frac{x^2}{16}+\frac{1}{2}+\frac{1}{x^2}=(a+b)^2[/math]
It's a perfect square which will take care of the square root.

Exactly! You will be integrating the absolute value of a simple function.
(x/4 - 1/x)^2 = (x^2)/16 - (1/2) + (1/(x^2)) so how do you think you can factor (x^2)/16 + (1/2) + (1/(x^2))?

 

[Steven G]

I would try substitution first. I have not completed this but I would start with:

x = (√8) * tan(Φ)

and see what do I get ........

Are you feeling ok today? I hope so.

 

[Evari5te]

Exactly! You will be integrating the absolute value of a simple function.
(x/4 - 1/x)^2 = (x^2)/16 - (1/2) + (1/(x^2)) so how do you think you can factor (x^2)/16 + (1/2) + (1/(x^2))?

Aahhhhh. Many thanks all and good spot - given the similarity between the resulting polynomial with the 1 added and the original (a - b)^2 it's an easy one to factor, just reverse the sign to (a +b), giving (x/4 + 1/x)^2.

Think that gets me away, so thanks again and very much appreciated!

 

[Steven G]

Just remember that the sqrt(x^2) is not x. After all, sqrt( (-4)^2 ) is not -4