What is the integral of (2xy)/(3x^2-y^2) ?

The integral of (2xy)/(3x^2-y^2) is 1/3 yln|-y^2+3x^2|+C

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integral of (2xy)/(3x^2-y^2)

\int \frac{2 x y}{3 x ^{2} - y ^{2}}

\frac{1}{3} y ln - y^{2} + 3 x^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{6 u} d u

= 2 y \cdot \int \frac{1}{6 u} d u

Take the constant out:

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

= 2 y \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 y \frac{1}{6} ln ∣ u ∣

Substitute back

u = - y^{2} + 3 x^{2}

= 2 y \frac{1}{6} ln - y^{2} + 3 x^{2}

Simplify

2 y \frac{1}{6} ln - y^{2} + 3 x^{2} : \frac{1}{3} y ln - y^{2} + 3 x^{2}

= \frac{1}{3} y ln - y^{2} + 3 x^{2}

Add a constant to the solution

= \frac{1}{3} y ln - y^{2} + 3 x^{2} + C

\int tan^{3} (4 x) sec (4 x) d x

f (x) = 9 x e^{x}, (0, 0)

f (x) = \frac{1}{5} x^{4} - 4 x^{2} + 1

\frac{d}{d x} (\frac{3 x + 2}{2 x + 1})

(x + 1) y^{^{'}} = x + 6

What is the integral of (2xy)/(3x^2-y^2) ?
The integral of (2xy)/(3x^2-y^2) is 1/3 yln|-y^2+3x^2|+C