What is the integral of (1-2x)/((x+1)(x-5)) ?

The integral of (1-2x)/((x+1)(x-5)) is -1/2 ln|x+1|-3/2 ln|x-5|+C

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integral of (1-2x)/((x+1)(x-5))

\int \frac{1 - 2 x}{( x + 1 ) ( x - 5 )} d x

- \frac{1}{2} ln ∣ x + 1 ∣ - \frac{3}{2} ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{1 - 2 x}{( x + 1 ) ( x - 5 )} d x

Take the partial fraction of

\frac{1 - 2 x}{( x + 1 ) ( x - 5 )} : - \frac{1}{2 ( x + 1 )} - \frac{3}{2 ( x - 5 )}

= \int - \frac{1}{2 ( x + 1 )} - \frac{3}{2 ( x - 5 )} 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}{2 ( x + 1 )} d x - \int \frac{3}{2 ( x - 5 )} d x

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

\int \frac{3}{2 ( x - 5 )} d x = \frac{3}{2} ln ∣ x - 5 ∣

= - \frac{1}{2} ln ∣ x + 1 ∣ - \frac{3}{2} ln ∣ x - 5 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x + 1 ∣ - \frac{3}{2} ln ∣ x - 5 ∣ + C

\frac{d}{d x} (x e^{x y} + 2 x y + \frac{1}{x})

y^{^{'}} - e^{y} - 2 x e^{y} = 0

y = sin (tan (7 x))

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

\int 56 cos (8 x - 5) d x

What is the integral of (1-2x)/((x+1)(x-5)) ?
The integral of (1-2x)/((x+1)(x-5)) is -1/2 ln|x+1|-3/2 ln|x-5|+C