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

The integral of (x^2-6)/((x+2)(2x-1)) is 1/2 x+2/5 ln|x+2|-23/20 ln|2x-1|+C

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

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

\frac{1}{2} x + \frac{2}{5} ln ∣ x + 2 ∣ - \frac{23}{20} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} - 6}{( x + 2 ) ( 2 x - 1 )} : \frac{1}{2} + \frac{2}{5 ( x + 2 )} - \frac{23}{10 ( 2 x - 1 )}

= \int \frac{1}{2} + \frac{2}{5 ( x + 2 )} - \frac{23}{10 ( 2 x - 1 )} 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} d x + \int \frac{2}{5 ( x + 2 )} d x - \int \frac{23}{10 ( 2 x - 1 )} d x

\int \frac{1}{2} d x = \frac{1}{2} x

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

\int \frac{23}{10 ( 2 x - 1 )} d x = \frac{23}{20} ln ∣ 2 x - 1 ∣

= \frac{1}{2} x + \frac{2}{5} ln ∣ x + 2 ∣ - \frac{23}{20} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{2} x + \frac{2}{5} ln ∣ x + 2 ∣ - \frac{23}{20} ln ∣ 2 x - 1 ∣ + C

y = x^{2}, y = 6 + x

\frac{d}{d x} (lo g_{e} (4 x))

\frac{d}{d x} ((8 (4 x)^{4} + 5)^{3})

\frac{\partial}{\partial y} (arctan (x) y)

4 - ln (x)

What is the integral of (x^2-6)/((x+2)(2x-1)) ?
The integral of (x^2-6)/((x+2)(2x-1)) is 1/2 x+2/5 ln|x+2|-23/20 ln|2x-1|+C