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

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

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

integral

\frac{3 x ^{3}}{( 1 - x )}

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

Solution steps

\int \frac{3 x ^{3}}{1 - x} d x

Take the constant out:

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

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

Long division

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

= 3 \cdot \int - 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

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

\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 ∣

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{1} + x^{2} \frac{x ^{2}}{2}

\frac{- y}{( 1 + y ^{2} )}

\frac{1}{x} (arctan (\frac{1}{x}))

- 4 e^{3} x

tan (\frac{x}{2})

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