What is the integral of (-x^2+5x-4)/((2-x)^3) ?

The integral of (-x^2+5x-4)/((2-x)^3) is ln|x-2|+1/(x-2)+1/((x-2)^2)+C

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integral of (-x^2+5x-4)/((2-x)^3)

\int \frac{- x ^{2} + 5 x - 4}{( 2 - x ) ^{3}} d x

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

Solution steps

\int \frac{- x ^{2} + 5 x - 4}{( 2 - x ) ^{3}} d x

Take the partial fraction of

\frac{- x ^{2} + 5 x - 4}{( 2 - x ) ^{3}} : \frac{1}{x - 2} - \frac{1}{( x - 2 ) ^{2}} - \frac{2}{( x - 2 ) ^{3}}

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

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

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

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (-x^2+5x-4)/((2-x)^3) ?
The integral of (-x^2+5x-4)/((2-x)^3) is ln|x-2|+1/(x-2)+1/((x-2)^2)+C