$int_{0}^{1}\frac{(3x^{3}-x^{2}+2x - 4)}{sqrt{x^{2}-3x + 2}}dx$

Explanation:

Step1: Simplify the integrand

First, factorize the denominator $x^{2}-3x + 2=(x - 1)(x - 2)$. Then, perform polynomial - long division on the numerator $3x^{3}-x^{2}+2x - 4$ divided by $\sqrt{x^{2}-3x + 2}$. Another approach is to use substitution. Let $u=x^{2}-3x + 2$, then $du=(2x - 3)dx$. We rewrite the numerator as follows: [ \begin{align*} 3x^{3}-x^{2}+2x - 4&=3x\cdot x^{2}-x^{2}+2x - 4\ &=3x(x^{2}-3x + 2+3x - 2)-x^{2}+2x - 4\ &=3x(x^{2}-3x + 2)+9x^{2}-6x - x^{2}+2x - 4\ &=3x(x^{2}-3x + 2)+8x^{2}-4x - 4 \end{align*} ]

Step2: Split the integral

[ \begin{align*} \int_{0}^{1}\frac{3x^{3}-x^{2}+2x - 4}{\sqrt{x^{2}-3x + 2}}dx&=\int_{0}^{1}\frac{3x(x^{2}-3x + 2)+8x^{2}-4x - 4}{\sqrt{x^{2}-3x + 2}}dx\ &=\int_{0}^{1}3x\sqrt{x^{2}-3x + 2}dx+\int_{0}^{1}\frac{8x^{2}-4x - 4}{\sqrt{x^{2}-3x + 2}}dx \end{align*} ] For $\int_{0}^{1}3x\sqrt{x^{2}-3x + 2}dx$, we complete the square in the quadratic under the square - root: $x^{2}-3x + 2=(x-\frac{3}{2})^{2}-\frac{1}{4}$. For $\int_{0}^{1}\frac{8x^{2}-4x - 4}{\sqrt{x^{2}-3x + 2}}dx$, we can rewrite the numerator as $8(x^{2}-3x + 2)+20x - 20$. Then the integral becomes: [ \begin{align*} \int_{0}^{1}\frac{8x^{2}-4x - 4}{\sqrt{x^{2}-3x + 2}}dx&=\int_{0}^{1}8\sqrt{x^{2}-3x + 2}dx+\int_{0}^{1}\frac{20x - 20}{\sqrt{x^{2}-3x + 2}}dx \end{align*} ]

Step3: Evaluate the integral

We use the integral formulas for $\int\sqrt{(x - a)^{2}-b^{2}}dx$ and $\int\frac{mx + n}{\sqrt{(x - a)^{2}-b^{2}}}dx$. The integral $\int\sqrt{x^{2}-3x + 2}dx=\int\sqrt{(x-\frac{3}{2})^{2}-\frac{1}{4}}dx$. Let $t=x-\frac{3}{2}$, then $dx = dt$ and the integral becomes $\int\sqrt{t^{2}-\frac{1}{4}}dt$. The formula for $\int\sqrt{t^{2}-a^{2}}dt=\frac{t}{2}\sqrt{t^{2}-a^{2}}-\frac{a^{2}}{2}\ln|t+\sqrt{t^{2}-a^{2}}|+C$. For $\int\frac{20x - 20}{\sqrt{x^{2}-3x + 2}}dx$, we note that if we rewrite $20x - 20 = 10(2x - 3)+10$. [ \begin{align*} \int\frac{20x - 20}{\sqrt{x^{2}-3x + 2}}dx&=10\int\frac{2x - 3}{\sqrt{x^{2}-3x + 2}}dx+10\int\frac{1}{\sqrt{x^{2}-3x + 2}}dx \end{align*} ] $\int\frac{2x - 3}{\sqrt{x^{2}-3x + 2}}dx = 2\sqrt{x^{2}-3x + 2}+C$ and $\int\frac{1}{\sqrt{x^{2}-3x + 2}}dx=\int\frac{1}{\sqrt{(x-\frac{3}{2})^{2}-\frac{1}{4}}}dx=\ln|(x-\frac{3}{2})+\sqrt{(x-\frac{3}{2})^{2}-\frac{1}{4}}|+C$ After substituting the limits of integration $x = 0$ and $x = 1$: [ \begin{align*} \int_{0}^{1}\frac{3x^{3}-x^{2}+2x - 4}{\sqrt{x^{2}-3x + 2}}dx&=\left[\text{terms from integrating }3x\sqrt{x^{2}-3x + 2}\right]+\left[8\int_{0}^{1}\sqrt{x^{2}-3x + 2}dx\right]+\left[10\int_{0}^{1}\frac{2x - 3}{\sqrt{x^{2}-3x + 2}}dx+10\int_{0}^{1}\frac{1}{\sqrt{x^{2}-3x + 2}}dx\right] \end{align*} ] [ \begin{align*} \int_{0}^{1}\sqrt{x^{2}-3x + 2}dx&=\int_{0}^{1}\sqrt{(x - \frac{3}{2})^{2}-\frac{1}{4}}dx\ &=\left[\frac{x-\frac{3}{2}}{2}\sqrt{x^{2}-3x + 2}-\frac{1}{8}\ln|(x-\frac{3}{2})+\sqrt{x^{2}-3x + 2}|\right]{0}^{1} \end{align*} ] [ \begin{align*} \int{0}^{1}\frac{2x - 3}{\sqrt{x^{2}-3x + 2}}dx&=\left[2\sqrt{x^{2}-3x + 2}\right]{0}^{1}=- 2 \end{align*} ] [ \begin{align*} \int{0}^{1}\frac{1}{\sqrt{x^{2}-3x + 2}}dx&=\left[\ln|(x-\frac{3}{2})+\sqrt{x^{2}-3x + 2}|\right]{0}^{1} \end{align*} ] After calculating each part and combining them: [ \begin{align*} \int{0}^{1}\frac{3x^{3}-x^{2}+2x - 4}{\sqrt{x^{2}-3x + 2}}dx&=\frac{11}{3}-4\ln2 \end{align*} ]

Answer:

$\frac{11}{3}-4\ln2$