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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

\frac{1}{x ( x - 3 ) ^{2}} : \frac{1}{9 x} - \frac{1}{9 ( x - 3 )} + \frac{1}{3 ( x - 3 ) ^{2}}

= 3 \cdot \int \frac{1}{9 x} - \frac{1}{9 ( x - 3 )} + \frac{1}{3 ( x - 3 ) ^{2}} 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

= 3 (\int \frac{1}{9 x} d x - \int \frac{1}{9 ( x - 3 )} d x + \int \frac{1}{3 ( x - 3 ) ^{2}} d x)

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int \frac{1}{9 ( x - 3 )} d x = \frac{1}{9} ln ∣ x - 3 ∣

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

= 3 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{3 ( x - 3 )})

Simplify

3 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{1}{3 ( x - 3 )}) : \frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x - 3 ∣ - \frac{1}{x - 3}

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

Add a constant to the solution

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

\int sec^{2} (y) d y

n = 1 \sum \infty (\frac{8}{9})^{n}

x \to 0 lim (\frac{7 x + x ^{2}}{x})

\int 30 (25 - x^{2}) d x

f (x) = 7 ln (x), at x = 3

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