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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{x ^{3} - 4}{x ^{2} + 2}

f (t) = \frac{1}{t}

x^{2} + 5 x

x ln (2 x) y^{^{'}} - y = 0

\frac{d}{d x} (c cos (x) + x^{2} sin (x))

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