What is the integral of (8x^3-16x^2+7)/(x^2-2x) ?

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

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integral of (8x^3-16x^2+7)/(x^2-2x)

\int \frac{8 x ^{3} - 16 x ^{2} + 7}{x ^{2} - 2 x} d x

4 x^{2} + \frac{7}{2} ln ∣ x - 2 ∣ - \frac{7}{2} ln ∣ x ∣ - 16 + C

Solution steps

\int \frac{8 x ^{3} - 16 x ^{2} + 7}{x ^{2} - 2 x} d x

Expand

\frac{8 x ^{3} - 16 x ^{2} + 7}{x ^{2} - 2 x} : \frac{8 x ^{3}}{x ^{2} - 2 x} - \frac{16 x ^{2}}{x ^{2} - 2 x} + \frac{7}{x ^{2} - 2 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{8 x ^{3}}{x ^{2} - 2 x} d x - \int \frac{16 x ^{2}}{x ^{2} - 2 x} d x + \int \frac{7}{x ^{2} - 2 x} d x

\int \frac{8 x ^{3}}{x ^{2} - 2 x} d x = 8 (\frac{( x - 2 ) ^{2}}{2} + 4 (x - 2) + 4 ln ∣ x - 2 ∣)

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

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

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

Simplify

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

= 4 x^{2} + \frac{7}{2} ln ∣ x - 2 ∣ - \frac{7}{2} ln ∣ x ∣ - 16

Add a constant to the solution

= 4 x^{2} + \frac{7}{2} ln ∣ x - 2 ∣ - \frac{7}{2} ln ∣ x ∣ - 16 + C

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

\int (x^{2} + 1)^{5} \cdot x d x

\frac{d ^{3}}{d x ^{3}} (ln (x))

\frac{d}{d x} (\frac{6}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}})

y = x^{- 8 c o s (x)}

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