What is the integral of ((6x^2+4))/((3y^2-4)) ?

The integral of ((6x^2+4))/((3y^2-4)) is 1/(-4+3y^2)(4x+2x^3)+C

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integral of ((6x^2+4))/((3y^2-4))

\int \frac{( 6 x ^{2} + 4 )}{( 3 y ^{2} - 4 )}

\frac{1}{- 4 + 3 y ^{2}} (4 x + 2 x^{3}) + C

Solution steps

\int \frac{6 x ^{2} + 4}{3 y ^{2} - 4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{- 4 + 3 y ^{2}} \cdot \int 4 + 6 x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{- 4 + 3 y ^{2}} (\int 4 d x + \int 6 x^{2} d x)

\int 4 d x = 4 x

\int 6 x^{2} d x = 2 x^{3}

= \frac{1}{- 4 + 3 y ^{2}} (4 x + 2 x^{3})

Add a constant to the solution

= \frac{1}{- 4 + 3 y ^{2}} (4 x + 2 x^{3}) + C

\frac{\partial}{\partial x} (4 x^{8} y^{6} + 5 x^{7} y^{5})

\int_{- \infty}^{8} e^{0.04 x} d x

(t + e^{t}) (3 - t)

x \to 0 lim (\frac{x}{cos ( x ) - 1})

x \to 5 lim (3 x - 3)

What is the integral of ((6x^2+4))/((3y^2-4)) ?
The integral of ((6x^2+4))/((3y^2-4)) is 1/(-4+3y^2)(4x+2x^3)+C