Find $\int \frac{(x^3 + x + 1)}{x^2 - 1} dx$

$\frac{x^2}{2} + \ln |x^2 - 1| + \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| + C$

The correct answer is Option (1) → $\frac{x^2}{2} + \ln |x^2 - 1| + \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| + C$

Since the degree of the numerator (3) is greater than the degree of the denominator (2), we will perform polynomial long division.

Thus, we can rewrite the integral as:

$\int \left( x + \frac{2x + 1}{x^2 - 1} \right) dx = \int x dx + \int \frac{2x + 1}{x^2 - 1} dx$

$= \int x dx + \int \frac{2x}{x^2 - 1} dx + \int \frac{1}{x^2 - 1} dx$

$= \frac{x^2}{2} + C_1 + \ln |x^2 - 1| + C_2 + I_3$

For $I_3 = \int \frac{1}{x^2 - 1} dx$ using partial fractions:

$\frac{1}{x^2 - 1} = \frac{1}{(x-1)(x+1)} = \frac{A}{(x-1)} + \frac{B}{(x+1)}$

$1 = A(x+1) + B(x-1)$

Put $x = 1$: $1 = 2A \Rightarrow A = \frac{1}{2}$

Put $x = -1$: $1 = -2B \Rightarrow B = -\frac{1}{2}$

Thus, we have:

$\frac{1}{x^2 - 1} = \frac{1/2}{(x-1)} - \frac{1/2}{(x+1)}$

Now integrating:

$\int \frac{1}{x^2 - 1} dx = \frac{1}{2} \ln |x - 1| - \frac{1}{2} \ln |x + 1| + C_3$

$= \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| + C_3$

Thus, $I =\int \frac{(x^3 + x + 1)}{x^2 - 1} dx = \frac{x^2}{2} + \ln |x^2 - 1| + \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| + C$ where $C=C_1+C_2+C_3$