Find $\int_{0}^{1} \frac{4x^{3}-6x^{2}+8x-3}{(x^{2}-x+1)^2}dx$.

Question

Find $\int_{0}^{1} \frac{4x^{3}-6x^{2}+8x-3}{(x^{2}-x+1)^2}dx$. I obtained that it is equal to $\int_{0}^{1} 2\frac{(x^2-x+1)'}{(x^{2}-x+1)^{n-1}}+\frac{(x^{2}-x+1)'}{(x^{2}-x+1)^{n}}dt$. I have problems with the interval of integration because, for $f(x)=x^{2}-x+1$, $f(1)=1$ and $f(0)=1$ and therefore the whole would be $\int_{1}^{1} 2\frac{1}{t^{n-1}}+\frac{1}{t^{n}}dt=0$ and I feel I make a mistake by these changes of integration edges.

Answer 1

$$\frac{4x^3-6x^2+8x-3}{(x^2-x+1)^2}=\frac{2(2x-1)(x^2-x+1)+2x-1}{(x^2-x+1)^2}$$

If$$f(x)=\frac{4x^3-6x^2+8x-3}{(x^2-x+1)^2}$$ then $$f(1-x)=-f(x)$$

So,answer is clearly ZERO

Answer 2

HInt:

One idea:

$$\dfrac{d\left(\dfrac{ax^2+bx+c}{x^2-x+1}\right)}{dx}=\dfrac{(2ax+b)(x^2-x+1)+(ax^2+bx+c)(2x-1)}{(x^2-x+1)^2}$$

Simplify the numerator & compare with $4x^3-6x^2+8x-3$