What is the integral of (20x^3-40x^2+7)/(x^2-2x) ?

The integral of (20x^3-40x^2+7)/(x^2-2x) is 10x^2-7(1/2 ln|x|-1/2 ln|x-2|)+C

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integral of (20x^3-40x^2+7)/(x^2-2x)

\int \frac{20 x ^{3} - 40 x ^{2} + 7}{x ^{2} - 2 x} d x

10 x^{2} - 7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

Solution steps

\int \frac{20 x ^{3} - 40 x ^{2} + 7}{x ^{2} - 2 x} d x

Long division

\frac{20 x ^{3} - 40 x ^{2} + 7}{x ^{2} - 2 x} : 20 x + \frac{7}{x ^{2} - 2 x}

= \int 20 x + \frac{7}{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 20 x d x + \int \frac{7}{x ^{2} - 2 x} d x

\int 20 x d x = 10 x^{2}

\int \frac{7}{x ^{2} - 2 x} d x = - 7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

= 10 x^{2} - 7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

Add a constant to the solution

= 10 x^{2} - 7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

f (x) = 3 e^{- x^{2}}

\int cos (ln (5 x)) d x

y^{^{'}} = \frac{11 x}{y}

y = cos (a^{6} + x^{6})

\int \frac{4}{15} x^{\frac{5}{2}} + 9 cos (x) + C x + C d x

What is the integral of (20x^3-40x^2+7)/(x^2-2x) ?
The integral of (20x^3-40x^2+7)/(x^2-2x) is 10x^2-7(1/2 ln|x|-1/2 ln|x-2|)+C