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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

- \frac{x ^{4}}{x - 1} - (- 4 (\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣)) : - \frac{x ^{4}}{x - 1} + 4 (\frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} + 3 (x - 1) + ln ∣ x - 1 ∣)

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

Add a constant to the solution

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

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