Question

1. $int\frac{dx}{x^{2}-x + 2}$

Explanation:

Step1: Complete the square for the denominator

First, complete the square for $x^{2}-x + 2$. We have $x^{2}-x+2=(x-\frac{1}{2})^{2}+2-\frac{1}{4}=(x - \frac{1}{2})^{2}+\frac{7}{4}$. So the integral becomes $2\int\frac{dx}{(x-\frac{1}{2})^{2}+\frac{7}{4}}$.

Step2: Use substitution

Let $u=x-\frac{1}{2}$, then $du = dx$. The integral is now $2\int\frac{du}{u^{2}+\frac{7}{4}}$.

Step3: Rewrite the denominator and use integral formula

We can rewrite $\frac{7}{4}$ as $(\frac{\sqrt{7}}{2})^{2}$. The integral formula for $\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan(\frac{u}{a})+C$. Here $a = \frac{\sqrt{7}}{2}$. So $2\int\frac{du}{u^{2}+\frac{7}{4}}=2\times\frac{2}{\sqrt{7}}\arctan(\frac{2u}{\sqrt{7}})+C$.

Step4: Substitute back $u$

Substitute $u=x - \frac{1}{2}$ back in. We get $\frac{4}{\sqrt{7}}\arctan(\frac{2x - 1}{\sqrt{7}})+C$.

Answer:

$\frac{4}{\sqrt{7}}\arctan(\frac{2x - 1}{\sqrt{7}})+C$