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

The integral of (x^2)/(2-x) is -1/2 (2-x)^2+8-4x-4ln|2-x|+C

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

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

- \frac{1}{2} (2 - x)^{2} + 8 - 4 x - 4 ln ∣ 2 - x ∣ + C

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{( - u + 2 ) ^{2}}{u} : u - 4 + \frac{4}{u}

= - \int u - 4 + \frac{4}{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 4 d u + \int \frac{4}{u} d u)

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

\int 4 d u = 4 u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= - (\frac{u ^{2}}{2} - 4 u + 4 ln ∣ u ∣)

Substitute back

u = 2 - x

= - (\frac{( 2 - x ) ^{2}}{2} - 4 (2 - x) + 4 ln ∣ 2 - x ∣)

Simplify

- (\frac{( 2 - x ) ^{2}}{2} - 4 (2 - x) + 4 ln ∣ 2 - x ∣) : - \frac{1}{2} (2 - x)^{2} + 8 - 4 x - 4 ln ∣ 2 - x ∣

= - \frac{1}{2} (2 - x)^{2} + 8 - 4 x - 4 ln ∣ 2 - x ∣

Add a constant to the solution

= - \frac{1}{2} (2 - x)^{2} + 8 - 4 x - 4 ln ∣ 2 - x ∣ + C

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What is the integral of (x^2)/(2-x) ?
The integral of (x^2)/(2-x) is -1/2 (2-x)^2+8-4x-4ln|2-x|+C