Evaluate $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$ This integral
$$\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
is fabricated based on an interesting point which I may post later.
The question is: How would you do it?

At first, we must check whether integrand function is bounded or not. [More specifically, if it is not bounded, we must approach this problem as improper integral..] Since both numerator and denominator are continuous at [0,1], we enough to check whether denominator has 0-value or not. Since $3\left(x-\dfrac{1}{2}\right)^{2}<3x^{2}-3x+1$, we can conclude that denominator does NOT have 0 value. So if we can get indefinite integral of integrand, then we can also calculate the definite integral.
Let's manipulate integrand function so that we can get indefinite integral. At first, I want to divide numerator into denominator.
$3x^{3}+3x^{2}-3x+1$
$=3(x^{3}+1)-3+(3x^{2}-3x+1)$
$=3(x+1)(x^{2}-x+1)-3+(3x^{2}-3x+1)$
$=(x+1)(3x^{2}-3x+3)-3+(3x^{2}-3x+1)$
$=(x+1)(3x^{2}-3x+1)+2(x+1)-3+(3x^{2}-3x+1)$
$=(x+2)(3x^{2}-3x+1)+(2x-1)$
From this calculation, we can manipulate integrand as $x+2+\dfrac{2x-1}{3(x^{2}-x)+1}$.
Indefinite integral of $x+2$ is $\dfrac{x^{2}}{2}+2x$, so we only need to compute indefinite integral of the last term. To integrate it, I will first substitute $x^{2}-x$ as u, so that $u'(x)=2x-1$. So it can be changed as $\dfrac{1}{3} \dfrac{1}{u+\dfrac{1}{3}}$ and its indefinite integral is $\dfrac{1}{3} \log{\left(u+\dfrac{1}{3}\right)}$. Since u(1)=u(0) in this situation, we only need to calculate $\left[\dfrac{x^{2}}{2}+2x \right]^{1}_{0}=\dfrac{5}{2}$.
