If \(I = \int \displaystyle\frac{x^4 + x^2 + 1}{x^2 - x + 1}\, dx = \alpha x + \beta x^2 + \gamma x^3 + \delta\), \(\delta\) is constant of integration, then \((\alpha + 2\beta + 3\gamma)\) equals
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
The numerator factorises as a difference of squares:
\(x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2\)
\(x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)\)
So the integrand reduces to \(x^2 + x + 1\):
\( \displaystyle I = \int (x^2 + x + 1)\,dx = \frac{x^3}{3} + \frac{x^2}{2} + x + \delta \)
Comparing with \(\alpha x + \beta x^2 + \gamma x^3 + \delta\):
\( \displaystyle \alpha = 1,\quad \beta = \frac{1}{2},\quad \gamma = \frac{1}{3} \)
\(\alpha + 2\beta + 3\gamma = 1 + 1 + 1 = 3\)