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

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

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

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

- ln ∣ 2 x + 1 ∣ + ln ∣ x - 2 ∣ + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

\int \frac{2}{5 ( 2 x + 1 )} d x = \frac{1}{5} ln ∣ 2 x + 1 ∣

\int \frac{1}{5 ( x - 2 )} d x = \frac{1}{5} ln ∣ x - 2 ∣

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

Simplify

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

= - ln ∣ 2 x + 1 ∣ + ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ 2 x + 1 ∣ + ln ∣ x - 2 ∣ + C

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