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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{( u + 1 ) ^{3}}{u} d u

= \int \frac{( u + 1 ) ^{3}}{u} d u

Expand

\frac{( u + 1 ) ^{3}}{u} : u^{2} + 3 u + 3 + \frac{1}{u}

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

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

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

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{3}}{3} + \frac{3 u ^{2}}{2} + 3 u + ln ∣ u ∣

Substitute back

u = x - 1

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

Add a constant to the solution

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

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

\frac{d y}{d x} = \frac{x - 1}{3 x ^{2} - 6 x - 1}

\int (ln (x)) 2 d x

\frac{ln ( x )}{ln ( 2 )}

\int \frac{4 x ^{2} + 9}{x ^{4}} d x

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