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

y^{^{''}} - 2 y^{^{'}} + 8 y = 320 t^{2}

\frac{\partial}{\partial z} (- 2 x z)

\frac{1}{3 x}

x \to 4 lim (\frac{x + 4}{x - 4})

\int \frac{t}{e ^{3 t}} d 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