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

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

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

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

Solution steps

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

Expand

\frac{2 - x ^{2}}{x - x ^{2}} : \frac{2}{x - x ^{2}} - \frac{x ^{2}}{x - 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{2}{x - x ^{2}} d x - \int \frac{x ^{2}}{x - x ^{2}} d x

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int 3 x^{5} - 6 x^{2} + x - 9 d x

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

\int \frac{2 x ^{2}}{a ^{2} + x ^{2}} d x

\int 2^{3 w} d w

\int (x^{- 3} - 6 x^{- 2} + 10) d x

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