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

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

integral

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

= \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

\frac{1}{( x ^{2} + a ^{2} ) ^{2}}

lo g_{10} (1 + x)

(x + y)^{2}

\frac{k \cdot x}{( 1 + x ^{2} )}

lo g_{10} (x + 2)

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