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We decompose the integrand into partial fractions. Let $$ \frac{1}{x(x^2 - 1)} = \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} $$ Multiplying both sides by $x(x-1)(x+1)$, we get $$ 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1) $$
To find $A$, let $x = 0$: $$ 1 = A(-1)(1) + B(0) + C(0) \implies A = -1 $$ To find $B$, let $x = 1$: $$ 1 = A(0) + B(1)(2) + C(0) \implies 1 = 2B \implies B = \frac{1}{2} $$ To find $C$, let $x = -1$: $$ 1 = A(0) + B(0) + C(-1)(-2) \implies 1 = 2C \implies C = \frac{1}{2} $$ Thus, we have $A = -1$, $B = \frac{1}{2}$, and $C = \frac{1}{2}$.
Now we can rewrite the integral as: $$ \int \frac{1}{x(x^2 - 1)} dx = \int \left( \frac{-1}{x} + \frac{1/2}{x-1} + \frac{1/2}{x+1} \right) dx $$
We integrate each term separately: $$ \int \frac{-1}{x} dx = -\ln|x| $$ $$ \int \frac{1/2}{x-1} dx = \frac{1}{2} \ln|x-1| $$ $$ \int \frac{1/2}{x+1} dx = \frac{1}{2} \ln|x+1| $$
Combining the results, we get: $$ \int \frac{1}{x(x^2 - 1)} dx = -\ln|x| + \frac{1}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| + C $$ We can simplify this further: $$ = -\ln|x| + \frac{1}{2} (\ln|x-1| + \ln|x+1|) + C $$ $$ = -\ln|x| + \frac{1}{2} \ln|(x-1)(x+1)| + C $$ $$ = -\ln|x| + \frac{1}{2} \ln|x^2 - 1| + C $$ $$ = \ln|x^{-1}| + \ln|\sqrt{x^2 - 1}| + C $$ $$ = \ln\left| \frac{\sqrt{x^2 - 1}}{x} \right| + C $$
Final Answer: $\ln\left| \frac{\sqrt{x^2 - 1}}{x} \right| + C$
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