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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{2 ( x - 1 )} - \frac{1}{2} arctan (x)

Add a constant to the solution

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

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\int \frac{5}{64 - 81 t ^{2}} d t

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