Integration and Partial Differentiation Problem

[Mastur]

Homework Statement

(A)
$\int{\frac{(v^2+2v+4)dv}{v^3+v^2+2v+4}}$

(B)
$\frac{\partial{M}}{\partial{y}}=(1-xy)^{-2}$

$\frac{\partial{N}}{\partial{x}}=y^2+x^2(1-xy)^{-2}$

Homework Equations

(A) How can I integrate this?

(B)After getting the partial derivatives, are they equal?

The Attempt at a Solution

(A)This is actually I stopped since I cannot integrate it. I tried factoring the denominator so I can use integration by partial fractions. Unfortunately, I cannot. Any hint in integrating it?

(B)
$\frac{\partial{M}}{\partial{y}}=\frac{2x}{1-xy}$

$\frac{\partial{N}}{\partial{x}}=\frac{2x+2x^2y-2x^3y^2}{(1-xy)^2}$

Can I still factor the second equation so I can get the same answer like the $\frac{\partial{M}}{\partial{y}}$?

 

[LCKurtz]

$\frac{\partial{M}}{\partial{y}}=(1-xy)^{-2}$

$\frac{\partial{N}}{\partial{x}}=y^2+x^2(1-xy)^{-2}$


(B)After getting the partial derivatives, are they equal?

Do they look equal if you try, for example, x = 0 and y = 0 in both?

 

[Mastur]

Do they look equal if you try, for example, x = 0 and y = 0 in both?

Yes, they will be equal after differentiating.

But that does not work all the time. I've tried checking my other answers in other problem, tried substituting x & y by 0, but not all are equal, but the equation is still exact after differentiating.

 

[LCKurtz]

We are obviously mis-communicating. Please state the original problem you are working on that involves the M and N in part B of your question. Presumably there is a differential equation involved.

Also you have said

$\frac{\partial{M}}{\partial{y}}=(1-xy)^{-2}$
in your question and

$\frac{\partial{M}}{\partial{y}}=\frac{2x}{1-xy}$

in your solution. They can't both be M_y

 

[Mastur]

Sorry for the confusion.

The equation is:

$M=(1-xy)^{-2}$

$N=y^2+x^2(1-xy)^{-2}$

And I got this function after differentiating.

$\frac{\partial{M}}{\partial{y}}=\frac{2x}{1-xy}$

$\frac{\partial{N}}{\partial{x}}=\frac{2x+2x^2y-2x^3y^2}{(1-xy)^2}$

My actual question is, is there any way to make the $\frac{\partial{M}}{\partial{y}} = \frac{\partial{N}}{\partial{x}}$??

 

[LCKurtz]

Sorry for the confusion.

The equation is:

$M=(1-xy)^{-2}$

$N=y^2+x^2(1-xy)^{-2}$

And I got this function after differentiating.

$\frac{\partial{M}}{\partial{y}}=\frac{2x}{1-xy}$

$\frac{\partial{N}}{\partial{x}}=\frac{2x+2x^2y-2x^3y^2}{(1-xy)^2}$

My actual question is, is there any way to make the $\frac{\partial{M}}{\partial{y}} = \frac{\partial{N}}{\partial{x}}$??

Yes, there is, but you need to calculate them correctly. Neither of your partials is correct. Given that both functions have a quantity to the -2 power, you would expect a denominator with exponent of 3 in your simplified answers. Perhaps if you write both M and N as quotients instead of using negative exponents and use the quotient rule, you might have better luck calculating the partials. Just guessing here, since you didn't show your work.