What is the integral of (16x^3-32x^2+7)/(x^2-2x) ?

The integral of (16x^3-32x^2+7)/(x^2-2x) is 8x^2-7(1/2 ln|x|-1/2 ln|x-2|)+C

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integral of (16x^3-32x^2+7)/(x^2-2x)

\int \frac{16 x ^{3} - 32 x ^{2} + 7}{x ^{2} - 2 x} d x

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

Solution steps

\int \frac{16 x ^{3} - 32 x ^{2} + 7}{x ^{2} - 2 x} d x

Long division

\frac{16 x ^{3} - 32 x ^{2} + 7}{x ^{2} - 2 x} : 16 x + \frac{7}{x ^{2} - 2 x}

= \int 16 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 16 x d x + \int \frac{7}{x ^{2} - 2 x} d x

\int 16 x d x = 8 x^{2}

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

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

Add a constant to the solution

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

4 δ (t) + 5 e^{t} - 5 e^{5 t}

y^{^{'}} = 0.06 (y - 65), y (0) = 190

f (t) = 10 cos (t)

x \to π lim (5 tan (x))

y^{^{'}} - e^{8 x + y} = 0

What is the integral of (16x^3-32x^2+7)/(x^2-2x) ?
The integral of (16x^3-32x^2+7)/(x^2-2x) is 8x^2-7(1/2 ln|x|-1/2 ln|x-2|)+C