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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2)/((1-x))

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

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

Solution steps

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

Apply u-substitution

= \int - \frac{( - u + 1 ) ^{2}}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{( - u + 1 ) ^{2}}{u} d u

Expand

\frac{( - u + 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

= - \int u - 2 + \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int u d u - \int 2 d u + \int \frac{1}{u} d u)

\int u d u = \frac{u ^{2}}{2}

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = 1 - x

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{6 x ^{4} + 4}{x ^{2}} d x

\int \frac{( \frac{b}{2} ) x}{2 a + b x} d x

\int 4 x + 12 d x

\int (cosh (x) + x^{7}) d x

\int x (x - x + 1) d x

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