\(\int \frac{x^4+x^2+1}{x^2-x+1} d x=\)

\(\int \frac{x^4+x^2+1}{x^2-x+1} d x=\)
  1. \(\left(\frac{1}{3}\right) x^3+\left(\frac{1}{2}\right) x^2+x+c\)
  2. \(\left(\frac{1}{3}\right) x^3-\left(\frac{1}{2}\right) x^2+x+c\)
  3. \(\left(\frac{1}{3}\right) x^3+\left(\frac{1}{2}\right) x^2-x+c\)
  4. \(\left(\frac{1}{3}\right) x^3-\left(\frac{1}{2}\right) x^2-x+c\)

Solution

We have, \(\begin{aligned} & I=\int \frac{x^4+x^2+1}{x^2-x+1} d x \\ & I=\int \frac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{x^2-x+1} d x \end{aligned}\) So, \(I=\int\left(x^2+x+1\right) d x=\frac{x^3}{3}+\frac{x^2}{2}+x+c\)

Asked in: AP EAMCET 2020 (17 Sep Shift 2)

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