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

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

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

\int \frac{x ^{4}}{1 - x} d x

- \frac{x ^{4}}{4} - \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - x - ln ∣ - x + 1 ∣ + C

Solution steps

\int \frac{x ^{4}}{1 - x} d x

Long division

\frac{x ^{4}}{1 - x} : - x^{3} - x^{2} - x - 1 + \frac{1}{- x + 1}

= \int - x^{3} - x^{2} - x - 1 + \frac{1}{- x + 1} d x

Apply the Sum Rule:

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

= - \int x^{3} d x - \int x^{2} d x - \int x d x - \int 1 d x + \int \frac{1}{- x + 1} d x

\int x^{3} d x = \frac{x ^{4}}{4}

\int x^{2} d x = \frac{x ^{3}}{3}

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

\int 1 d x = x

\int \frac{1}{- x + 1} d x = - ln ∣ - x + 1 ∣

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (y - z)

\frac{d}{d u} (e^{u})

l a pl a ce t r an s f or m 5 t - \frac{π}{3}

\frac{d y}{d x} = 4 x (x - 2)

n = 1 \sum \infty - \frac{1}{n}

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