1. The problem is to find the integral $$\int \frac{1}{t^2 - a^2} \, dt$$ where $a$ is a constant. 2. Recognize that the denominator can be factored using the difference of squares formula: $$t^2 - a^2 = (t - a)(t + a)$$ 3. Use partial fraction decomposition to express the integrand: $$\frac{1}{t^2 - a^2} = \frac{1}{(t - a)(t + a)} = \frac{A}{t - a} + \frac{B}{t + a}$$ 4. Multiply both sides by $(t - a)(t + a)$ to clear denominators: $$1 = A(t + a) + B(t - a)$$ 5. Expand the right side: $$1 = A t + A a + B t - B a = (A + B) t + (A a - B a)$$ 6. Equate coefficients of like terms: - Coefficient of $t$: $A + B = 0$ - Constant term: $A a - B a = 1$ 7. From $A + B = 0$, we get $B = -A$. 8. Substitute $B = -A$ into the constant term equation: $$A a - (-A) a = A a + A a = 2 A a = 1$$ 9. Solve for $A$: $$A = \frac{1}{2 a}$$ 10. Then $B = -\frac{1}{2 a}$. 11. Rewrite the integral using partial fractions: $$\int \frac{1}{t^2 - a^2} \, dt = \int \left( \frac{1}{2 a (t - a)} - \frac{1}{2 a (t + a)} \right) dt$$ 12. Integrate term by term: $$= \frac{1}{2 a} \int \frac{1}{t - a} dt - \frac{1}{2 a} \int \frac{1}{t + a} dt$$ 13. The integrals are natural logarithms: $$= \frac{1}{2 a} \ln|t - a| - \frac{1}{2 a} \ln|t + a| + C$$ 14. Combine the logarithms: $$= \frac{1}{2 a} \ln \left| \frac{t - a}{t + a} \right| + C$$ **Final answer:** $$\int \frac{1}{t^2 - a^2} \, dt = \frac{1}{2 a} \ln \left| \frac{t - a}{t + a} \right| + C$$