Integration Question (1 Viewer)

[joesmith1975]

If the gradient of a curve is given by dy/dx = x/(x^2-1)^2 and the curve passes throught (2,1), find the equation of the curve.

Thanks in advance.

 

[pLuvia]

∫ x/(x²-1)² dx
Let u = x²-1 du=2x dx

.: 1/2 ∫ 2x/(x²-1)² dx

1/2 ∫ du/u²
= 1/2 [ -1/u ]
= -1/2(x²-1) + C

Since it passes through (2,1) sub into the equation to find C
1 = -1/2(4-1) + C
C = 7/6

.: y = -1/2(x²-1) + 7/6

Hope that helped

 

[Trev]

dy/dx = x/(x²-1)²; through (2,1).
∴ y = ∫ [x/(x²-1)²].dx
y = -1/2*1/(x²-1) + C
Sub in coordinates to find C and you will get the equ. of the curve.

 

[joesmith1975]

The answer is y= -1/2(x^2-1) -5/6 not 7/6 :S

 

[pLuvia]

Are you sure?

 

[joesmith1975]

thats what it says in the back on the book

 

[pLuvia]

Well sometimes it is wrong

Looking at trev's and mine both are the same and both produce the same answer

 

[Slidey]

I just lost my working. Anyway, suffice to say that pLuvia is correct. The book forgot to multiply by negative one when integrating the equivalent of 1/x^2.

 

[joesmith1975]

o.k no worries any ways, thanks

 

[joesmith1975]

sorry guys i put goes through point (2,1) but i read the text book and its (2,-1)

I've worked through it and have completed it, thanks for your help.

 

[Slidey]

That's also a valid explanation for the discrepancy in answers.