What is the integral of 9/(x(x-9)^2) ?

The integral of 9/(x(x-9)^2) is 1/9 ln|x|-1/9 ln|x-9|-1/(x-9)+C

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integral of 9/(x(x-9)^2)

\int \frac{9}{x ( x - 9 ) ^{2}} d x

\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 9 ∣ - \frac{1}{x - 9} + C

Solution steps

\int \frac{9}{x ( x - 9 ) ^{2}} d x

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{x ( x - 9 ) ^{2}} : \frac{1}{81 x} - \frac{1}{81 ( x - 9 )} + \frac{1}{9 ( x - 9 ) ^{2}}

= 9 \cdot \int \frac{1}{81 x} - \frac{1}{81 ( x - 9 )} + \frac{1}{9 ( x - 9 ) ^{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

= 9 (\int \frac{1}{81 x} d x - \int \frac{1}{81 ( x - 9 )} d x + \int \frac{1}{9 ( x - 9 ) ^{2}} d x)

\int \frac{1}{81 x} d x = \frac{1}{81} ln ∣ x ∣

\int \frac{1}{81 ( x - 9 )} d x = \frac{1}{81} ln ∣ x - 9 ∣

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

= 9 (\frac{1}{81} ln ∣ x ∣ - \frac{1}{81} ln ∣ x - 9 ∣ - \frac{1}{9 ( x - 9 )})

Simplify

9 (\frac{1}{81} ln ∣ x ∣ - \frac{1}{81} ln ∣ x - 9 ∣ - \frac{1}{9 ( x - 9 )}) : \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 9 ∣ - \frac{1}{x - 9}

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 9 ∣ - \frac{1}{x - 9}

Add a constant to the solution

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 9 ∣ - \frac{1}{x - 9} + C

\int (tan^{2} (x) 4) d x

\frac{d}{d x} (e^{2 x^{6}} - 1)

\frac{\partial}{\partial y} (ln (x^{6} + y^{7}))

x \to 6 - lim (cos (\frac{1}{x - 6}))

\frac{\partial}{\partial y} (x^{2} + y^{2} + x y)

What is the integral of 9/(x(x-9)^2) ?
The integral of 9/(x(x-9)^2) is 1/9 ln|x|-1/9 ln|x-9|-1/(x-9)+C