What is the integral of (16x^3-32x^2+5)/(x^2-2x) ?

The integral of (16x^3-32x^2+5)/(x^2-2x) is 8x^2-5(1/2 ln|x|-1/2 ln|x-2|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (16x^3-32x^2+5)/(x^2-2x)

\int \frac{16 x ^{3} - 32 x ^{2} + 5}{x ^{2} - 2 x} d x

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

Solution steps

\int \frac{16 x ^{3} - 32 x ^{2} + 5}{x ^{2} - 2 x} d x

Long division

\frac{16 x ^{3} - 32 x ^{2} + 5}{x ^{2} - 2 x} : 16 x + \frac{5}{x ^{2} - 2 x}

= \int 16 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 16 x d x + \int \frac{5}{x ^{2} - 2 x} d x

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

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

= 8 x^{2} - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

Add a constant to the solution

= 8 x^{2} - 5 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C

\int 8^{3 x + 5} d x

in v erse l a pl a ce \frac{2}{( s ^{2} + 4 ) ^{2}}

p a r i t y x csc (x)

\frac{\partial}{\partial y} (- arctan (\frac{x}{y}))

\int (e^{x} + 1) d x

What is the integral of (16x^3-32x^2+5)/(x^2-2x) ?
The integral of (16x^3-32x^2+5)/(x^2-2x) is 8x^2-5(1/2 ln|x|-1/2 ln|x-2|)+C