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

The integral of (2x^4)/((x-1)^3) is -(x^4)/((x-1)^2)+2x^2+8x-4/(x-1)+12ln|x-1|-10+C

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

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

- \frac{x ^{4}}{( x - 1 ) ^{2}} + 2 x^{2} + 8 x - \frac{4}{x - 1} + 12 ln ∣ x - 1 ∣ - 10 + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{2 x ^{3}}{( x - 1 ) ^{2}} d x = - x^{2} - 4 x + \frac{2}{x - 1} - 6 ln ∣ x - 1 ∣ + 5

= 2 (- \frac{x ^{4}}{2 ( x - 1 ) ^{2}} - (- x^{2} - 4 x + \frac{2}{x - 1} - 6 ln ∣ x - 1 ∣ + 5))

Simplify

2 (- \frac{x ^{4}}{2 ( x - 1 ) ^{2}} - (- x^{2} - 4 x + \frac{2}{x - 1} - 6 ln ∣ x - 1 ∣ + 5)) : - \frac{x ^{4}}{( x - 1 ) ^{2}} + 2 x^{2} + 8 x - \frac{4}{x - 1} + 12 ln ∣ x - 1 ∣ - 10

= - \frac{x ^{4}}{( x - 1 ) ^{2}} + 2 x^{2} + 8 x - \frac{4}{x - 1} + 12 ln ∣ x - 1 ∣ - 10

Add a constant to the solution

= - \frac{x ^{4}}{( x - 1 ) ^{2}} + 2 x^{2} + 8 x - \frac{4}{x - 1} + 12 ln ∣ x - 1 ∣ - 10 + C

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What is the integral of (2x^4)/((x-1)^3) ?
The integral of (2x^4)/((x-1)^3) is -(x^4)/((x-1)^2)+2x^2+8x-4/(x-1)+12ln|x-1|-10+C