Let $f : R \rightarrow R$ be a function as $f(x) = (x - 1)(x + 2)(x - 3)(x - 6) - 100$. If $g(x)$ is a polynomial of degree $\leq 3$ such that $\displaystyle \int \frac{g(x)}{f(x)} dx$ does not contain any logarithm function and $g(-2) = 10$. Then

$\displaystyle \int \frac{g(x)}{f(x)} dx$, equals

Here, $f(x) = (x - 1)(x + 2)(x - 3)(x - 6) - 100$

$= (x^2 + 4x + 3) (x^2 - 4x - 12) - 100$

$= (x^2 - 4x)^2 - 9(x^2 - 4x) - 136$

$= (x^2 - 4x + 8)(x^2 - 4x + 17)$

$\displaystyle \int \frac{g(x)}{f(x)} = \frac{g(x)}{(x^2 - 4x - 17)(x^2 - 4x + 8)}$

$= \frac{Ax + B}{x^2 - 4x - 17} + \frac{Cx + D}{x^2 - 4x + 8}$

Clearly, $A, B$ and $C$ must be zero.

$\therefore \frac{g(x)}{(x^2 - 4x - 17)(x^2 - 4x + 8)} = \frac{D}{x^2 - 4x + 8}$

$\therefore$ $g(x) = D (x^2 - 4x - 17)$

$g(-2) = D (4 + 8 - 17) = -10$ [given]

$\Rightarrow$ $\frac{g(x)}{f(x)} = \frac{2 (x^2 - 4x - 17)}{(x^2 - 4x - 17)(x^2 - 4x + 8)} = \frac{2}{x^2 - 4x + 8}$

$\therefore$ $\displaystyle \int \frac{g(x)}{f(x)} dx = \displaystyle \int \frac{2}{x^2 - 4x + 8}dx = 2 \displaystyle \int \frac{dx}{(x - 2)^2 + (2)^2}$

$= 2. \frac{1}{2} \tan^{-1} \left ( \frac{x - 2}{2} \right ) + C = \tan^{-1} \left ( \frac{x - 2}{2} \right ) + C$