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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

\int \frac{1}{y - 2} d y = ln ∣ y - 2 ∣

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

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

Add a constant to the solution

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

\int \frac{e ^{x} + e ^{- x}}{e ^{x} - e ^{- x}} d x

x^{2} y^{^{''}} + 2 x y^{^{'}} - 12 y = 0, y (1) = 0, y^{^{'}} (1) = - 2

n = 1 \sum \infty \frac{1}{n ^{0}}

x \frac{d y}{d x} + 4 y = x^{3} - x

(1.3) (4.2)

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