What is the integral of (4x^3-8x^2+5)/(x^2-2x) ?

The integral of (4x^3-8x^2+5)/(x^2-2x) is 2x^2-5(1/2 ln|x|-1/2 ln|x-2|)+C

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integral of (4x^3-8x^2+5)/(x^2-2x)

\int \frac{4 x ^{3} - 8 x ^{2} + 5}{x ^{2} - 2 x} d x

2 x^{2} - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{4 x ^{3} - 8 x ^{2} + 5}{x ^{2} - 2 x} d x

Long division

\frac{4 x ^{3} - 8 x ^{2} + 5}{x ^{2} - 2 x} : 4 x + \frac{5}{x ^{2} - 2 x}

= \int 4 x + \frac{5}{x ^{2} - 2 x} 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 4 x d x + \int \frac{5}{x ^{2} - 2 x} d x

\int 4 x d x = 2 x^{2}

\int \frac{5}{x ^{2} - 2 x} d x = - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

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

Add a constant to the solution

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

\frac{d}{d y} (x + \frac{1}{2} ln (x^{2} + y^{2}))

\int 5 tan^{2} (x) + tan^{4} (x) d x

\int e^{(- x^{2})} d x

y = x^{2}, y = 2 - x, x = 0

f (x) = ln (\frac{e ^{x}}{1 + e ^{x}})

What is the integral of (4x^3-8x^2+5)/(x^2-2x) ?
The integral of (4x^3-8x^2+5)/(x^2-2x) is 2x^2-5(1/2 ln|x|-1/2 ln|x-2|)+C