If $\int\frac{x^3 + x + 1}{x^4 + x^2 + 1}dx = A_1\ln(x^2 + x + 1) + A_2\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + A_3\tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right) + A_4\tan^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right) + c$ then the value of $(A_1 + A_2 + A_3 + A_4)$ is
Step-by-Step Solution
Key Concept: Decompose a rational function by strategically splitting the numerator to match partial denominators and apply substitution techniques.
The integral $I = \int \frac{(x^2 + x + 1)dx}{(x^2 + x + 1)(x^2 - x + 1)}$ is decomposed by splitting the numerator into $(x^2 - x + 1) + (x^2 + x + 1) - x$ in the standard form. This yields three integrals: $I_1$ (with numerator $xdx$), $I_2$ (with numerator $dx$), and $I_3$ (with numerator $xdx$). For $I_3$, substituting $x^2 = t$ transforms it to a standard arctangent form: $I_3 = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x^2 + 1}{\sqrt{3}}\right)$.
Correct Answer: I need to find the coefficients $A_1, A_2, A_3, A_4$ in the given integral decomposition.
First, let me factor the denominator:
$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$