evaluate: ∫(x² + 1)/(x⁴ - x² + 1) dx

Explanation:

Step1: Divide numerator and denominator by $x^{2}$

Divide the integrand $\frac{x^{2}+1}{x^{4}-x^{2}+1}$ by $x^{2}$ to get $\int\frac{1 + \frac{1}{x^{2}}}{x^{2}-1+\frac{1}{x^{2}}}dx$.

Step2: Rearrange the denominator

We know that $x^{2}+\frac{1}{x^{2}}=(x - \frac{1}{x})^{2}+2$. So the integrand becomes $\int\frac{1+\frac{1}{x^{2}}}{(x-\frac{1}{x})^{2}+1}dx$.

Step3: Use substitution

Let $t=x-\frac{1}{x}$, then $dt=(1 + \frac{1}{x^{2}})dx$. The integral $\int\frac{1+\frac{1}{x^{2}}}{(x-\frac{1}{x})^{2}+1}dx$ becomes $\int\frac{dt}{t^{2}+1}$.

Step4: Integrate

The integral of $\frac{1}{t^{2}+1}$ with respect to $t$ is $\arctan(t)+C$.

Step5: Substitute back

Substitute $t = x-\frac{1}{x}$ back into the result. We get $\arctan(x-\frac{1}{x})+C$.

Answer:

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