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

The integral of (x^2)/(2(2-x)) is -1/4 (x^2+4x+8ln|-x+2|-12)+C

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

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

- \frac{1}{4} (x^{2} + 4 x + 8 ln ∣ - x + 2 ∣ - 12) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= \frac{1}{2} \cdot \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

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

Expand

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

= \frac{1}{2} (- \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

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

Substitute back

u = 2 - x

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

Simplify

\frac{1}{2} (- (\frac{( 2 - x ) ^{2}}{2} - 4 (2 - x) + 4 ln ∣ 2 - x ∣)) : - \frac{1}{4} (x^{2} + 4 x + 8 ln ∣ - x + 2 ∣ - 12)

= - \frac{1}{4} (x^{2} + 4 x + 8 ln ∣ - x + 2 ∣ - 12)

Add a constant to the solution

= - \frac{1}{4} (x^{2} + 4 x + 8 ln ∣ - x + 2 ∣ - 12) + C

x \to 0 lim (\frac{- 1}{14 x})

d er i v a t i v e y = 6 x^{2} sin (x) tan (x)

x \to 2 lim (\frac{2 x + 4}{x - 2})

\int 4 x^{2} sin (x) d x

t an g e n t y = 2 x, (8, 4)

What is the integral of (x^2)/(2(2-x)) ?
The integral of (x^2)/(2(2-x)) is -1/4 (x^2+4x+8ln|-x+2|-12)+C