What is the integral of (4/(x^3)+2)/(1/(x^2)-x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4/(x^3)+2)/(1/(x^2)-x)

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

\frac{4}{3} (ln x^{3} - ln x^{3} - 1) - \frac{2}{3} ln 1 - x^{3} + C

Solution steps

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

Expand

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

\int \frac{\frac{4}{x ^{3}}}{\frac{1}{x ^{2}} - x} d x = \frac{4}{3} (ln x^{3} - ln x^{3} - 1)

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

= \frac{4}{3} (ln x^{3} - ln x^{3} - 1) - \frac{2}{3} ln 1 - x^{3}

Add a constant to the solution

= \frac{4}{3} (ln x^{3} - ln x^{3} - 1) - \frac{2}{3} ln 1 - x^{3} + C

\frac{d y}{d x} + 2 y = 1

f (x) = 6 x arcsin (x)

\frac{\partial}{\partial x} (arcsin (4 x yz))

f (x) = 18 x + 18, (1, 6)

(2 x y + 2 y^{2}) d x + (x^{2} + 4 x y) d y = 0

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