What is the integral of (4x-5)/((x-2)(x+3)) ?

The integral of (4x-5)/((x-2)(x+3)) is 17/5 ln|x+3|+3/5 ln|x-2|+C

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integral of (4x-5)/((x-2)(x+3))

\int \frac{4 x - 5}{( x - 2 ) ( x + 3 )} d x

\frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{4 x - 5}{( x - 2 ) ( x + 3 )} d x

Apply u-substitution

: \int \frac{4 u - 17}{u ( u - 5 )} d u

= \int \frac{4 u - 17}{u ( u - 5 )} d u

Take the partial fraction of

\frac{4 u - 17}{u ( u - 5 )} : \frac{17}{5 u} + \frac{3}{5 ( u - 5 )}

= \int \frac{17}{5 u} + \frac{3}{5 ( u - 5 )} d u

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{17}{5 u} d u + \int \frac{3}{5 ( u - 5 )} d u

\int \frac{17}{5 u} d u = \frac{17}{5} ln ∣ u ∣

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

= \frac{17}{5} ln ∣ u ∣ + \frac{3}{5} ln ∣ u - 5 ∣

Substitute back

u = x + 3

= \frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x + 3 - 5 ∣

Simplify

\frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x + 3 - 5 ∣ : \frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x - 2 ∣

= \frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{17}{5} ln ∣ x + 3 ∣ + \frac{3}{5} ln ∣ x - 2 ∣ + C

\frac{d}{d x} (\frac{1}{sin ( x - sin ( x ) )})

y = \frac{8 x ^{2} + 4 x + 2}{x}

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

\frac{d y}{d x} = 14 x^{- 8}

\frac{d}{d x} (x^{3} - 3 x^{2} - 9 x + 15)

What is the integral of (4x-5)/((x-2)(x+3)) ?
The integral of (4x-5)/((x-2)(x+3)) is 17/5 ln|x+3|+3/5 ln|x-2|+C