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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} + 5 (tan (5 x)) y = 4 cos (5 x)

x \to 0 + lim (\frac{x ^{2}}{4} - \frac{2}{x})

\frac{d}{d x} (x (x + 4))

\int sec^{2} (5 x t a) n^{3} 5 x d x

\int cos (x) cos (5 x) d x

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