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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

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

= - \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣ - 2 ln ∣ u ∣ + ln ∣ u + 1 ∣ + ln ∣ u - 1 ∣

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

\frac{cot ( x )}{2}

(a - x^{2})^{\frac{1}{2}}

sin^{3} (x) cos^{3} (x)

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

sin^{2} (x) cos^{3} (x)

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