What is the integral of (-2x^4)/((1-x)^3) ?

The integral of (-2x^4)/((1-x)^3) is -(x^4)/((1-x)^2)+(4x^3)/(1-x)+6x^2+12x+12ln|-x+1|-18+C

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integral of (-2x^4)/((1-x)^3)

\int \frac{- 2 x ^{4}}{( 1 - x ) ^{3}} d x

- \frac{x ^{4}}{( 1 - x ) ^{2}} + \frac{4 x ^{3}}{1 - x} + 6 x^{2} + 12 x + 12 ln ∣ - x + 1 ∣ - 18 + C

Solution steps

\int \frac{- 2 x ^{4}}{( 1 - x ) ^{3}} d x

Take the constant out:

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

= - 2 \cdot \int \frac{x ^{4}}{( 1 - x ) ^{3}} d x

Apply Integration By Parts

: \frac{x ^{4}}{2 ( 1 - x ) ^{2}} - \int \frac{2 x ^{3}}{( 1 - x ) ^{2}} d x

= - 2 (\frac{x ^{4}}{2 ( 1 - x ) ^{2}} - \int \frac{2 x ^{3}}{( 1 - x ) ^{2}} d x)

\int \frac{2 x ^{3}}{( 1 - x ) ^{2}} d x = 2 (\frac{x ^{3}}{1 - x} + 3 (\frac{1}{2} x^{2} + x + ln ∣ - x + 1 ∣ + \frac{1}{2} - 2))

= - 2 (\frac{x ^{4}}{2 ( 1 - x ) ^{2}} - 2 (\frac{x ^{3}}{1 - x} + 3 (\frac{1}{2} x^{2} + x + ln ∣ - x + 1 ∣ + \frac{1}{2} - 2)))

Simplify

- 2 (\frac{x ^{4}}{2 ( 1 - x ) ^{2}} - 2 (\frac{x ^{3}}{1 - x} + 3 (\frac{1}{2} x^{2} + x + ln ∣ - x + 1 ∣ + \frac{1}{2} - 2))) : - \frac{x ^{4}}{( 1 - x ) ^{2}} + \frac{4 x ^{3}}{1 - x} + 6 x^{2} + 12 x + 12 ln ∣ - x + 1 ∣ - 18

= - \frac{x ^{4}}{( 1 - x ) ^{2}} + \frac{4 x ^{3}}{1 - x} + 6 x^{2} + 12 x + 12 ln ∣ - x + 1 ∣ - 18

Add a constant to the solution

= - \frac{x ^{4}}{( 1 - x ) ^{2}} + \frac{4 x ^{3}}{1 - x} + 6 x^{2} + 12 x + 12 ln ∣ - x + 1 ∣ - 18 + C

sin (x^{3})

f (t) = 2 t

\frac{d}{d z} (\frac{sin ( z )}{( z + 1 ) ^{2}})

x \to 1 lim (\frac{x - 1}{x ^{6} - 1})

y^{^{''}} + 4 y = t^{2}

What is the integral of (-2x^4)/((1-x)^3) ?
The integral of (-2x^4)/((1-x)^3) is -(x^4)/((1-x)^2)+(4x^3)/(1-x)+6x^2+12x+12ln|-x+1|-18+C