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

The integral of (x+1)/((x+2)(x-3)) is 1/5 ln|x+2|+4/5 ln|x-3|+C

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

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

\frac{1}{5} ln ∣ x + 2 ∣ + \frac{4}{5} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x + 1}{( x + 2 ) ( x - 3 )} : \frac{1}{5 ( x + 2 )} + \frac{4}{5 ( x - 3 )}

= \int \frac{1}{5 ( x + 2 )} + \frac{4}{5 ( x - 3 )} 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}{5 ( x + 2 )} d x + \int \frac{4}{5 ( x - 3 )} d x

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

\int \frac{4}{5 ( x - 3 )} d x = \frac{4}{5} ln ∣ x - 3 ∣

= \frac{1}{5} ln ∣ x + 2 ∣ + \frac{4}{5} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ x + 2 ∣ + \frac{4}{5} ln ∣ x - 3 ∣ + C

\int (tan (2 x) + cot (2 x))^{2} d x

\frac{6 x}{( 3 - 2 x ) ^{3}}

\frac{d}{d x} ((6 x + 5)^{2})

y^{^{''}} + 16 y = 24 sin (4 t)

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

What is the integral of (x+1)/((x+2)(x-3)) ?
The integral of (x+1)/((x+2)(x-3)) is 1/5 ln|x+2|+4/5 ln|x-3|+C