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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{3 x ^{2}}{8} - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

Solution steps

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

Long division

\frac{3 x ^{3} - 3 x + 4}{4 x ^{2} - 4} : \frac{3 x}{4} + \frac{1}{x ^{2} - 1}

= \int \frac{3 x}{4} + \frac{1}{x ^{2} - 1} 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 \frac{3 x}{4} d x + \int \frac{1}{x ^{2} - 1} d x

\int \frac{3 x}{4} d x = \frac{3 x ^{2}}{8}

\int \frac{1}{x ^{2} - 1} d x = - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣

= \frac{3 x ^{2}}{8} - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{3 x ^{2}}{8} - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + C

\frac{d}{d x} (e^{x^{\frac{1}{3}}} + (e^{x})^{\frac{1}{3}})

y = 1 1^{x}

\frac{d}{d x} (\frac{5}{x ^{\frac{1}{3}}})

\frac{\partial}{\partial x} (2 x e^{x^{2} - y^{2}})

x \to 0 lim (x (1 - \frac{1}{x}))

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