What is the integral of 1/((2x^2+6x-8)) ?

The integral of 1/((2x^2+6x-8)) is 1/10 ln|x-1|-1/10 ln|x+4|+C

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integral of 1/((2x^2+6x-8))

integral

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

\frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{1}{2 x ^{2} + 6 x - 8} d x

Take the partial fraction of

\frac{1}{2 x ^{2} + 6 x - 8} : \frac{1}{10 ( x - 1 )} - \frac{1}{10 ( x + 4 )}

= \int \frac{1}{10 ( x - 1 )} - \frac{1}{10 ( x + 4 )} 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

= \int \frac{1}{10 ( x - 1 )} d x - \int \frac{1}{10 ( x + 4 )} d x

\int \frac{1}{10 ( x - 1 )} d x = \frac{1}{10} ln ∣ x - 1 ∣

\int \frac{1}{10 ( x + 4 )} d x = \frac{1}{10} ln ∣ x + 4 ∣

= \frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln ∣ x + 4 ∣ + C

x (x - 1) (x + 1)

\frac{1}{( x ^{2} + 0.75 )}

f (x) = (1 + cos (6 x))^{\frac{1}{2}}

2 (sin (x))^{3} + sin (x) + tan^{5} (x)

\frac{x ^{2}}{( 1 + x ^{2} )}

What is the integral of 1/((2x^2+6x-8)) ?
The integral of 1/((2x^2+6x-8)) is 1/10 ln|x-1|-1/10 ln|x+4|+C