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

The integral of 2/(2x^2+4x-6) is 2(1/8 ln|x-1|-1/8 ln|x+3|)+C

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

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

2 (\frac{1}{8} ln ∣ x - 1 ∣ - \frac{1}{8} ln ∣ x + 3 ∣) + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{2 x ^{2} + 4 x - 6} : \frac{1}{8 ( x - 1 )} - \frac{1}{8 ( x + 3 )}

= 2 \cdot \int \frac{1}{8 ( x - 1 )} - \frac{1}{8 ( x + 3 )} 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

= 2 (\int \frac{1}{8 ( x - 1 )} d x - \int \frac{1}{8 ( x + 3 )} d x)

\int \frac{1}{8 ( x - 1 )} d x = \frac{1}{8} ln ∣ x - 1 ∣

\int \frac{1}{8 ( x + 3 )} d x = \frac{1}{8} ln ∣ x + 3 ∣

= 2 (\frac{1}{8} ln ∣ x - 1 ∣ - \frac{1}{8} ln ∣ x + 3 ∣)

Add a constant to the solution

= 2 (\frac{1}{8} ln ∣ x - 1 ∣ - \frac{1}{8} ln ∣ x + 3 ∣) + C

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

\int \frac{2 x + 3}{9 - 4 x ^{2}} d x

\frac{\partial}{\partial x} (5 x^{4} y^{6} + 3 x^{5} y^{8})

x \to 0 lim (\frac{( 6 ^{x} - 4 ^{x} )}{x})

\frac{9 x}{9 x - 5}

What is the integral of 2/(2x^2+4x-6) ?
The integral of 2/(2x^2+4x-6) is 2(1/8 ln|x-1|-1/8 ln|x+3|)+C