What is the integral of ((4x-11))/((x^3-9x^2)) ?

The integral of ((4x-11))/((x^3-9x^2)) is -25/81 ln|x|-(11)/(9x)+25/81 ln|x-9|+C

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integral of ((4x-11))/((x^3-9x^2))

\int \frac{( 4 x - 11 )}{( x ^{3} - 9 x ^{2} )} d x

- \frac{25}{81} ln ∣ x ∣ - \frac{11}{9 x} + \frac{25}{81} ln ∣ x - 9 ∣ + C

Solution steps

\int \frac{4 x - 11}{x ^{3} - 9 x ^{2}} d x

Take the partial fraction of

\frac{4 x - 11}{x ^{3} - 9 x ^{2}} : - \frac{25}{81 x} + \frac{11}{9 x ^{2}} + \frac{25}{81 ( x - 9 )}

= \int - \frac{25}{81 x} + \frac{11}{9 x ^{2}} + \frac{25}{81 ( x - 9 )} 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

= - \int \frac{25}{81 x} d x + \int \frac{11}{9 x ^{2}} d x + \int \frac{25}{81 ( x - 9 )} d x

\int \frac{25}{81 x} d x = \frac{25}{81} ln ∣ x ∣

\int \frac{11}{9 x ^{2}} d x = - \frac{11}{9 x}

\int \frac{25}{81 ( x - 9 )} d x = \frac{25}{81} ln ∣ x - 9 ∣

= - \frac{25}{81} ln ∣ x ∣ - \frac{11}{9 x} + \frac{25}{81} ln ∣ x - 9 ∣

Add a constant to the solution

= - \frac{25}{81} ln ∣ x ∣ - \frac{11}{9 x} + \frac{25}{81} ln ∣ x - 9 ∣ + C

\frac{d}{d x} (cos (wt) + sin (wt))

\int \frac{1}{2} e^{2 x} + \frac{1}{2} d x

x^{2} + 1

n = 5 \sum \infty \frac{1}{n + 25}

\frac{d}{d t} (\frac{6}{t})

What is the integral of ((4x-11))/((x^3-9x^2)) ?
The integral of ((4x-11))/((x^3-9x^2)) is -25/81 ln|x|-(11)/(9x)+25/81 ln|x-9|+C