What is the integral of (5x^2-3x+2)/(x^3-2x^2) ?

The integral of (5x^2-3x+2)/(x^3-2x^2) is ln|x|+4ln|x-2|+1/x+C

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integral of (5x^2-3x+2)/(x^3-2x^2)

\int \frac{5 x ^{2} - 3 x + 2}{x ^{3} - 2 x ^{2}} d x

ln ∣ x ∣ + 4 ln ∣ x - 2 ∣ + \frac{1}{x} + C

Solution steps

\int \frac{5 x ^{2} - 3 x + 2}{x ^{3} - 2 x ^{2}} d x

Expand

\frac{5 x ^{2} - 3 x + 2}{x ^{3} - 2 x ^{2}} : \frac{5 x ^{2}}{x ^{3} - 2 x ^{2}} - \frac{3 x}{x ^{3} - 2 x ^{2}} + \frac{2}{x ^{3} - 2 x ^{2}}

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{5 x ^{2}}{x ^{3} - 2 x ^{2}} d x - \int \frac{3 x}{x ^{3} - 2 x ^{2}} d x + \int \frac{2}{x ^{3} - 2 x ^{2}} d x

\int \frac{5 x ^{2}}{x ^{3} - 2 x ^{2}} d x = 5 ln ∣ x - 2 ∣

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

\int \frac{2}{x ^{3} - 2 x ^{2}} d x = - \frac{1}{2} ln ∣ x ∣ + \frac{1}{x} + \frac{1}{2} ln ∣ x - 2 ∣

= 5 ln ∣ x - 2 ∣ - 3 (- \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣) - \frac{1}{2} ln ∣ x ∣ + \frac{1}{x} + \frac{1}{2} ln ∣ x - 2 ∣

Simplify

5 ln ∣ x - 2 ∣ - 3 (- \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣) - \frac{1}{2} ln ∣ x ∣ + \frac{1}{x} + \frac{1}{2} ln ∣ x - 2 ∣ : ln ∣ x ∣ + 4 ln ∣ x - 2 ∣ + \frac{1}{x}

= ln ∣ x ∣ + 4 ln ∣ x - 2 ∣ + \frac{1}{x}

Add a constant to the solution

= ln ∣ x ∣ + 4 ln ∣ x - 2 ∣ + \frac{1}{x} + C

x \to 3 lim (x^{2} - 5)

y^{^{'}} = \frac{y + x ^{2} - y ^{2}}{x}

\int_{- 2.5}^{4.5} x^{3} d x

\frac{d}{d x} (a cos (x))

x \to - 1 lim (5 - 3 x)

What is the integral of (5x^2-3x+2)/(x^3-2x^2) ?
The integral of (5x^2-3x+2)/(x^3-2x^2) is ln|x|+4ln|x-2|+1/x+C