1. **Problem Statement:** Determine which of the following definite integrals can be evaluated over the given limits: (a) $$\int_{-1}^1 \frac{x+1}{x-1} \, dx$$ (b) $$\int_{-1}^1 \sin^{-1}(x) \, dx$$ (c) $$\int_{-1}^1 \frac{1}{x} \, dx$$ (d) $$\int_0^{\pi} \tan(x) \, dx$$ 2. **Key Concepts:** - A definite integral is evaluable if the integrand is continuous on the interval. - If the integrand has discontinuities or singularities (like division by zero) inside the interval, the integral may be improper or undefined. 3. **Analyze each integral:** (a) $$\frac{x+1}{x-1}$$ has a denominator zero at $$x=1$$, which is the upper limit. The function is not defined at $$x=1$$, so the integral is improper and needs special treatment. (b) $$\sin^{-1}(x)$$ (inverse sine) is continuous and defined for $$x \in [-1,1]$$. So this integral is evaluable. (c) $$\frac{1}{x}$$ has a discontinuity at $$x=0$$, which lies inside the interval $$[-1,1]$$. So this integral is improper and not directly evaluable without considering limits. (d) $$\tan(x)$$ has vertical asymptotes at $$x=\frac{\pi}{2}$$, which lies inside the interval $$[0,\pi]$$. So this integral is improper and not directly evaluable. 4. **Conclusion:** - (a) Improper integral due to discontinuity at upper limit. - (b) Proper integral, evaluable. - (c) Improper integral due to discontinuity inside interval. - (d) Improper integral due to discontinuity inside interval. **Final answer:** Only integral (b) $$\int_{-1}^1 \sin^{-1}(x) \, dx$$ can be evaluated directly. $$\boxed{\int_{-1}^1 \sin^{-1}(x) \, dx \text{ is evaluable}}$$