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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int 1 d x = x

\int \frac{3}{x - 1} d x = 3 ln ∣ x - 1 ∣

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

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

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

Add a constant to the solution

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

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

5 x^{3} - x^{2} y + y^{3} + 1 = 0, (1, - 2)

y = t^{3} - 4 t - 3, (2, - 3)

x \to 2 + lim (\frac{3}{( x - 2 ) ^{3}})

\int (y e^{y}) d y

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