1. **Problem Statement:** Determine which of the following definite integrals can be evaluated: (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) $$\int_{\pi}^{0} \tan(x) \, dx$$ 2. **Key Points:** - A definite integral is evaluable if the integrand is continuous on the interval of integration. - If the integrand has discontinuities or singularities within the interval, the integral may be improper or undefined. 3. **Analyze each integral:** (a) $$\frac{x+1}{x-1}$$ has a discontinuity at $$x=1$$ because the denominator is zero there. The interval is from 1 to -1, which includes $$x=1$$ as an endpoint. Since the integrand is not defined at $$x=1$$, the integral is improper but can be evaluated as an improper integral if limits are taken carefully. (b) $$\sin^{-1}(x)$$ (inverse sine) is defined and continuous for $$x \in [-1,1]$$. The interval is from 1 to -1, which is within the domain. This integral can be evaluated. (c) $$\frac{1}{x}$$ has a discontinuity at $$x=0$$. The interval is from 1 to -1, which includes $$x=0$$ inside the interval. Therefore, the integral is improper and cannot be evaluated as a standard definite integral without considering improper integral techniques. (d) $$\tan(x)$$ has discontinuities at $$x=\frac{\pi}{2}$$, which lies between 0 and $$\pi$$. The interval is from $$\pi$$ to 0, so it includes $$\frac{\pi}{2}$$. Thus, the integral is improper and cannot be evaluated as a standard definite integral. 4. **Summary:** - (a) Improper integral due to discontinuity at endpoint, can be evaluated as improper integral. - (b) Continuous integrand on interval, integral can be evaluated. - (c) Discontinuity inside interval, improper integral, not straightforward. - (d) Discontinuity inside interval, improper integral, not straightforward. **Final answer:** Only integral (b) can be evaluated directly as a proper definite integral.