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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

\int \frac{2 x + 1}{x ^{2} + 1} d x = ln x^{2} + 1 + arctan (x)

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

Add a constant to the solution

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

x \to 0 lim (x + 4)

x \to 0 lim (\frac{e ^{\frac{- 1}{x}}}{x})

\int z^{2} e^{- 3 z^{3}} d z

\int_{- 5}^{5} (25 - x^{2})^{2} d x

\frac{d}{d x} (cos (5 x) \cdot 5)

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