Question

$\int \frac{x^4+x^2+1}{x^2-x+1}dx=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$\int \frac{x^4+x^2+1}{x^2-x+1}dx$

$=\int \frac{x^4+2x^2+1-x^2}{x^2-x+1}dx$

$=\int \frac{{\left(x^2+1\right)}^2-x^2}{x^2-x+1}dx$

$=\int \frac{\left(x^2+1+x\right)\left(x^2+1-x\right)}{x^2-x+1}dx$

$=\int \left(x^2+1+x\right)dx$

$=\int x^{2}dx+\int xdx+\int dx$

$=\frac{x^3}{3}+\frac{x^2}{2}+x+C$


$\int \frac{x^4+x^2+1}{x^2-x+1}dx=\underline{\frac{x^3}{3}+\frac{x^2}{2}+x+C}.$