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

The integral of (x^3-1)/(x^3(x-2)^2) is -3/16 ln|x|+1/(4x)+1/(8x^2)+3/16 ln|x-2|-7/(8(x-2))+C

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integral of (x^3-1)/(x^3(x-2)^2)

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

- \frac{3}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{8 x ^{2}} + \frac{3}{16} ln ∣ x - 2 ∣ - \frac{7}{8 ( x - 2 )} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{3} - 1}{x ^{3} ( x - 2 ) ^{2}} : - \frac{3}{16 x} - \frac{1}{4 x ^{2}} - \frac{1}{4 x ^{3}} + \frac{3}{16 ( x - 2 )} + \frac{7}{8 ( x - 2 ) ^{2}}

= \int - \frac{3}{16 x} - \frac{1}{4 x ^{2}} - \frac{1}{4 x ^{3}} + \frac{3}{16 ( x - 2 )} + \frac{7}{8 ( x - 2 ) ^{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

= - \int \frac{3}{16 x} d x - \int \frac{1}{4 x ^{2}} d x - \int \frac{1}{4 x ^{3}} d x + \int \frac{3}{16 ( x - 2 )} d x + \int \frac{7}{8 ( x - 2 ) ^{2}} d x

\int \frac{3}{16 x} d x = \frac{3}{16} ln ∣ x ∣

\int \frac{1}{4 x ^{2}} d x = - \frac{1}{4 x}

\int \frac{1}{4 x ^{3}} d x = - \frac{1}{8 x ^{2}}

\int \frac{3}{16 ( x - 2 )} d x = \frac{3}{16} ln ∣ x - 2 ∣

\int \frac{7}{8 ( x - 2 ) ^{2}} d x = - \frac{7}{8 ( x - 2 )}

= - \frac{3}{16} ln ∣ x ∣ - (- \frac{1}{4 x}) - (- \frac{1}{8 x ^{2}}) + \frac{3}{16} ln ∣ x - 2 ∣ - \frac{7}{8 ( x - 2 )}

Simplify

= - \frac{3}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{8 x ^{2}} + \frac{3}{16} ln ∣ x - 2 ∣ - \frac{7}{8 ( x - 2 )}

Add a constant to the solution

= - \frac{3}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{8 x ^{2}} + \frac{3}{16} ln ∣ x - 2 ∣ - \frac{7}{8 ( x - 2 )} + C

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What is the integral of (x^3-1)/(x^3(x-2)^2) ?
The integral of (x^3-1)/(x^3(x-2)^2) is -3/16 ln|x|+1/(4x)+1/(8x^2)+3/16 ln|x-2|-7/(8(x-2))+C