To solve the integral [tex]\[ \int \frac{x^2 - x - 2}{1 - x^2} \, dx, \][/tex] we can follow these steps:
1. Simplify the integrand:
Notice that the denominator [tex]\( 1 - x^2 \)[/tex] can be factored as [tex]\( (1 - x)(1 + x) \)[/tex].
The integrand becomes:
[tex]\[ \frac{x^2 - x - 2}{1 - x^2} = \frac{x^2 - x - 2}{(1 - x)(1 + x)}. \][/tex]
Simplifying the numerator [tex]\(x^2 - x - 2\)[/tex], we can factor it as [tex]\( (x - 2)(x + 1) \)[/tex].
So, the integrand can be rewritten as:
[tex]\[ \frac{(x - 2)(x + 1)}{(1 - x)(1 + x)}. \][/tex]
2. Simplify further:
Notice that [tex]\( (1 - x) \)[/tex] can be written as [tex]\(-(x - 1)\)[/tex]:
[tex]\[ \frac{(x - 2)(x + 1)}{(1 - x)(1 + x)} = \frac{(x - 2)(x + 1)}{-[-(x - 1)](x + 1)} = -\frac{(x - 2)(x + 1)}{(x - 1)(x + 1)}. \][/tex]
The [tex]\((x + 1)\)[/tex] terms cancel out:
[tex]\[ -\frac{(x - 2)}{(x - 1)}. \][/tex]
So the integrand simplifies to:
[tex]\[ \int -\frac{x - 2}{x - 1} \, dx \][/tex]
3. Split the fraction:
We can split the fraction in the integrand:
[tex]\[ -\frac{x - 2}{x - 1} = -\left( \frac{x - 1}{x - 1} + \frac{-1}{x - 1} \right) = -\left( 1 + \frac{-1}{x - 1} \right). \][/tex]
Simplifying further:
[tex]\[ -\left( 1 + \frac{-1}{x - 1} \right) = -1 - \frac{-1}{x - 1} = -1 + \frac{1}{x - 1}. \][/tex]
So the integrand now is:
[tex]\[ \int \left( -1 + \frac{1}{x - 1} \right) \, dx. \][/tex]
4. Integrate term-by-term:
We can integrate each term individually:
[tex]\[ \int \left( -1 + \frac{1}{x - 1} \right) \, dx = \int -1 \, dx + \int \frac{1}{x - 1} \, dx. \][/tex]
The integral of [tex]\(-1\)[/tex] is:
[tex]\[ \int -1 \, dx = -x. \][/tex]
The integral of [tex]\(\frac{1}{x - 1}\)[/tex] is:
[tex]\[ \int \frac{1}{x - 1} \, dx = \log |x - 1|. \][/tex]
Putting everything together:
[tex]\[ -x + \log |x - 1| + C, \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
Hence, the integral [tex]\[ \int \frac{x^2 - x - 2}{1 - x^2} \, dx \][/tex] evaluates to:
[tex]\[ -x + \log |x - 1| + C. \][/tex]
