What is the integral of (6x^3)/(x-1) ?

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

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integral of (6x^3)/(x-1)

\int \frac{6 x ^{3}}{x - 1} d x

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

Solution steps

\int \frac{6 x ^{3}}{x - 1} d x

Take the constant out:

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

= 6 \cdot \int \frac{x ^{3}}{x - 1} d x

Apply u-substitution

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

= 6 \cdot \int \frac{( u + 1 ) ^{3}}{u} d u

Expand

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

= 6 \cdot \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

= 6 (\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 ∣

= 6 (\frac{u ^{3}}{3} + \frac{3 u ^{2}}{2} + 3 u + ln ∣ u ∣)

Substitute back

u = x - 1

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

Add a constant to the solution

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

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