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

The integral of (x+1)/(x(x-2)(x-3)) is 1/6 ln|x|-3/2 ln|x-2|+4/3 ln|x-3|+C

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

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

\frac{1}{6} ln ∣ x ∣ - \frac{3}{2} ln ∣ x - 2 ∣ + \frac{4}{3} ln ∣ x - 3 ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{1}{x ( x - 2 ) ( x - 3 )} d x = \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{3} ln ∣ x - 3 ∣

= - ln ∣ x - 2 ∣ + ln ∣ x - 3 ∣ + \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{3} ln ∣ x - 3 ∣

Simplify

- ln ∣ x - 2 ∣ + ln ∣ x - 3 ∣ + \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣ + \frac{1}{3} ln ∣ x - 3 ∣ : \frac{1}{6} ln ∣ x ∣ - \frac{3}{2} ln ∣ x - 2 ∣ + \frac{4}{3} ln ∣ x - 3 ∣

= \frac{1}{6} ln ∣ x ∣ - \frac{3}{2} ln ∣ x - 2 ∣ + \frac{4}{3} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ x ∣ - \frac{3}{2} ln ∣ x - 2 ∣ + \frac{4}{3} ln ∣ x - 3 ∣ + C

\frac{d}{d y} (\frac{x ^{2}}{2} - x y)

4 x^{2} - 2 x + 6

\int \frac{2 + 3 sin ^{2} ( x )}{sin ^{2} ( x )} d x

y^{^{'}} + \frac{4}{x ^{2} - 1} y = 0

\int 4 t^{2} d t

What is the integral of (x+1)/(x(x-2)(x-3)) ?
The integral of (x+1)/(x(x-2)(x-3)) is 1/6 ln|x|-3/2 ln|x-2|+4/3 ln|x-3|+C