$\int \frac{x^4+x^2+1}{x^2-x+1} \mathrm{~d} x$ is equal to

Question

Aicharya · Accepted Answer

$\begin{aligned} \mathrm{I} & =\int \frac{x^4+x^2+1}{x^2-x+1} \ & =\int\left(\frac{x^2-x+1+x^4+x}{x^2-x+1}ight) \mathrm{d} x \ & =\int \frac{x^2-x+1}{x^2-x+1} \mathrm{~d} x+\int \frac{\left(x^4+xight)}{x^2-x+1} \mathrm{~d} x \ & =\int 1 \mathrm{~d} x+\int \frac{x\left(x^3+1ight)}{x^2-x+1} \mathrm{~d} x \ & =x+\int \frac{x(x+1)\left(x^2-x+1ight)}{\left(x^2-x+1ight)} \mathrm{d} x \ & =x+\int\left(x^2+xight) \mathrm{d} x=x+\frac{x^3}{3}+\frac{x^2}{2}+\mathrm{c}\end{aligned}$

$\int \frac{x^4+x^2+1}{x^2-x+1} \mathrm{~d} x$ is equal to

Solution