How do you integrate #y=1/(-2+2x^2)# using the quotient rule?

Question

Luke Alvarez · Accepted Answer

I assume mean differentiate, not integrate, as the quotient rule is an differentiation tool!

# dy/dx = - (4x)/(-2+2x^2)^2 # You need to use the quotient rule; # d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 # So if # y = 1/(-2+2x^2) # we have; # dy/dx = {(-2+2x^2)(d/dx1) - (1)(d/dx(-2+2x^2))}/(-2+2x^2)^2 # # :. dy/dx = {0 - (4x)}/(-2+2x^2)^2 # # :. dy/dx = - (4x)/(-2+2x^2)^2 #

Christopher Allers · Answer

undefined To integrate $ y = \frac{1}{-2 + 2x^2} $ using the quotient rule, follow these steps:

1. Rewrite the integrand as a fraction: $ y = \frac{1}{-2 + 2x^2} $.
2. Decompose the denominator into partial fractions.
3. Integrate each partial fraction separately.

For the denominator $ -2 + 2x^2 $, rewrite it as $ -2(1 - x^2) $.

So, $ -2 + 2x^2 = -2(1 - x^2) $.

Now, rewrite the fraction as partial fractions:

$$ y = \frac{1}{-2(1 - x^2)} $$

$$ y = \frac{A}{1 - x^2} $$

Now, integrate each partial fraction separately. Since $ \int \frac{1}{1 - x^2} \, dx $ can be represented as $ \int \frac{1}{1 - u^2} \, du $, we can use a standard integral formula to integrate it.

$$ \int \frac{1}{1 - x^2} \, dx = \frac{1}{2} \ln \left| \frac{1 + x}{1 - x} \right| + C $$

Now, replace $ u $ with $ x $ in the final result.