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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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