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

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

integral

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

- \frac{1}{2} (- 1 + x^{2} + ln 1 - x^{2}) - (- \frac{1}{2} ln 1 - x^{2}) - (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣) : \frac{1}{2} - \frac{1}{2} x^{2} - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

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

Add a constant to the solution

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

x^{3} - 3 x^{2} + 5 x + 2

(e^{- t}) \cdot (\frac{1}{t})

x^{2} + a^{2}

(\frac{1}{x - 1})^{\frac{1}{2}}

\frac{10}{x ^{3}}

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