What is the integral of 1/((2x+3)(x-1)) ?

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

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integral of 1/((2x+3)(x-1))

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

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

Solution steps

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

Apply u-substitution

= \int \frac{1}{u ( u - 5 )} d u

Take the partial fraction of

\frac{1}{u ( u - 5 )} : - \frac{1}{5 u} + \frac{1}{5 ( u - 5 )}

= \int - \frac{1}{5 u} + \frac{1}{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{1}{5 u} d u + \int \frac{1}{5 ( u - 5 )} d u

\int \frac{1}{5 u} d u = \frac{1}{5} ln ∣ u ∣

\int \frac{1}{5 ( u - 5 )} d u = \frac{1}{5} ln ∣ u - 5 ∣

= - \frac{1}{5} ln ∣ u ∣ + \frac{1}{5} ln ∣ u - 5 ∣

Substitute back

u = 2 x + 3

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

Simplify

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

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

Add a constant to the solution

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

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