What is the integral of (x^4-6x^3+12x^2+6)/(x^3-6x^2+12x-8) ?

The integral of (x^4-6x^3+12x^2+6)/(x^3-6x^2+12x-8) is (x^2)/2-8/(x-2)-(11)/((x-2)^2)+C

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integral of (x^4-6x^3+12x^2+6)/(x^3-6x^2+12x-8)

\int \frac{x ^{4} - 6 x ^{3} + 12 x ^{2} + 6}{x ^{3} - 6 x ^{2} + 12 x - 8} d x

\frac{x ^{2}}{2} - \frac{8}{x - 2} - \frac{11}{( x - 2 ) ^{2}} + C

Solution steps

\int \frac{x ^{4} - 6 x ^{3} + 12 x ^{2} + 6}{x ^{3} - 6 x ^{2} + 12 x - 8} d x

Take the partial fraction of

\frac{x ^{4} - 6 x ^{3} + 12 x ^{2} + 6}{x ^{3} - 6 x ^{2} + 12 x - 8} : x + \frac{8}{( x - 2 ) ^{2}} + \frac{22}{( x - 2 ) ^{3}}

= \int x + \frac{8}{( x - 2 ) ^{2}} + \frac{22}{( 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 x d x + \int \frac{8}{( x - 2 ) ^{2}} d x + \int \frac{22}{( x - 2 ) ^{3}} d x

\int x d x = \frac{x ^{2}}{2}

\int \frac{8}{( x - 2 ) ^{2}} d x = - \frac{8}{x - 2}

\int \frac{22}{( x - 2 ) ^{3}} d x = - \frac{11}{( x - 2 ) ^{2}}

= \frac{x ^{2}}{2} - \frac{8}{x - 2} - \frac{11}{( x - 2 ) ^{2}}

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{8}{x - 2} - \frac{11}{( x - 2 ) ^{2}} + C

\frac{d}{d x} (\frac{( x + 4 ) ( x ^{2} - 4 x + 16 )}{x ^{3}})

5 x sin (x)

\int e^{- x} - 1 d x

\frac{\partial}{\partial x} (- x^{2})

\frac{\partial ^{2}}{\partial r \partial θ} (r (θ) \cdot sin (θ))

What is the integral of (x^4-6x^3+12x^2+6)/(x^3-6x^2+12x-8) ?
The integral of (x^4-6x^3+12x^2+6)/(x^3-6x^2+12x-8) is (x^2)/2-8/(x-2)-(11)/((x-2)^2)+C