Key Concepts and Formulas

• Integration by Substitution: $\int f(g(x))g'(x) dx = F(g(x)) + C$, where $F'(u) = f(u)$.
• Trigonometric Identities: $\sin^2(x) + \cos^2(x) = 1$, $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$.
• Standard Integrals: $\int \frac{dx}{\sqrt{x^2-1}} = \ln|x + \sqrt{x^2-1}| + C$, $\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}(x) + C$.

Step-by-Step Solution

Step 1: Rewrite the integral. We are given the integral: $I = \int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} dx$ Multiply the numerator and denominator of the fraction inside the square root by $(1+x)$ to simplify: $I = \int \frac{1}{x} \sqrt{\frac{(1-x)(1+x)}{(1+x)^2}} dx = \int \frac{1}{x} \frac{\sqrt{1-x^2}}{1+x} dx$ This form doesn't immediately lead to a simpler integration. Instead, let's multiply by $\sqrt{1-x}$ in numerator and denominator inside the square root: $I = \int \frac{1}{x} \sqrt{\frac{1-x}{1+x} \cdot \frac{1-x}{1-x}} dx = \int \frac{1}{x} \frac{1-x}{\sqrt{1-x^2}} dx$ $I = \int \frac{1-x}{x\sqrt{1-x^2}} dx = \int \frac{1}{x\sqrt{1-x^2}} dx - \int \frac{x}{x\sqrt{1-x^2}} dx$ $I = \int \frac{1}{x\sqrt{1-x^2}} dx - \int \frac{1}{\sqrt{1-x^2}} dx$

Step 2: Evaluate the first integral using substitution. Let $x = \frac{1}{t}$. Then, $dx = -\frac{1}{t^2} dt$. Substituting these into the first integral: $\int \frac{1}{x\sqrt{1-x^2}} dx = \int \frac{1}{\frac{1}{t}\sqrt{1-\frac{1}{t^2}}} \left(-\frac{1}{t^2}\right) dt = \int \frac{-1/t^2}{\frac{1}{t}\frac{\sqrt{t^2-1}}{t}} dt = \int \frac{-1}{\sqrt{t^2-1}} dt$ $= -\int \frac{1}{\sqrt{t^2-1}} dt = -\ln|t + \sqrt{t^2-1}| + C_1$ Substituting back $t = \frac{1}{x}$: $= -\ln\left|\frac{1}{x} + \sqrt{\frac{1}{x^2}-1}\right| + C_1$

Step 3: Evaluate the second integral. The second integral is a standard integral: $\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1}(x) + C_2$

Step 4: Combine the results. Substituting the results from Step 2 and Step 3 back into the expression for $I$: $I = -\ln\left|\frac{1}{x} + \sqrt{\frac{1}{x^2}-1}\right| - \sin^{-1}(x) + C$ Since $\sin^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x)$, we have: $I = -\ln\left|\frac{1}{x} + \sqrt{\frac{1}{x^2}-1}\right| - \left(\frac{\pi}{2} - \cos^{-1}(x)\right) + C$ $I = -\ln\left|\frac{1}{x} + \sqrt{\frac{1}{x^2}-1}\right| + \cos^{-1}(x) - \frac{\pi}{2} + C$ $I = \cos^{-1}(x) - \ln\left|\frac{1}{x} + \sqrt{\frac{1}{x^2}-1}\right| + C'$ where $C' = C - \frac{\pi}{2}$.

Step 5: Determine g(x). We are given that $I = g(x) + c$, so $g(x) = \cos^{-1}(x) - \ln\left|\frac{1}{x} + \sqrt{\frac{1}{x^2}-1}\right|$

Step 6: Use the condition g(1) = 0 to verify. $g(1) = \cos^{-1}(1) - \ln\left|\frac{1}{1} + \sqrt{\frac{1}{1^2}-1}\right| = 0 - \ln|1+0| = 0 - \ln(1) = 0$ This confirms our expression for $g(x)$.

Step 7: Calculate g(1/2). $g\left(\frac{1}{2}\right) = \cos^{-1}\left(\frac{1}{2}\right) - \ln\left|\frac{1}{\frac{1}{2}} + \sqrt{\frac{1}{(\frac{1}{2})^2}-1}\right| = \cos^{-1}\left(\frac{1}{2}\right) - \ln|2 + \sqrt{4-1}|$ $g\left(\frac{1}{2}\right) = \frac{\pi}{3} - \ln(2+\sqrt{3})$ Since $\ln(a) = -\ln(\frac{1}{a})$, $g\left(\frac{1}{2}\right) = \frac{\pi}{3} + \ln\left(\frac{1}{2+\sqrt{3}}\right) = \frac{\pi}{3} + \ln\left(\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}\right)$ $g\left(\frac{1}{2}\right) = \frac{\pi}{3} + \ln\left(\frac{2-\sqrt{3}}{4-3}\right) = \frac{\pi}{3} + \ln(2-\sqrt{3})$ Multiply by $\frac{2+\sqrt{3}}{2+\sqrt{3}}$ to rationalize the denominator of $\frac{\sqrt{3}-1}{\sqrt{3}+1}$. We can rewrite $2 - \sqrt{3}$ as $\frac{(2 - \sqrt{3})(2 + \sqrt{3})}{2 + \sqrt{3}} = \frac{4 - 3}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}}$. We want to express it in the form $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$. $\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$. Therefore, $g(\frac{1}{2}) = \frac{\pi}{3} + \ln\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$.

Common Mistakes & Tips

• Sign Errors: Be extra careful with negative signs during substitution and simplification.
• Constant of Integration: Remember to include the constant of integration (+C) after each indefinite integral. However, we can combine them into a single constant at the end.
• Simplifying Logarithms: Use logarithm properties to simplify the final expression.

Summary

We evaluated the given indefinite integral by using substitution and standard integral formulas. We then applied the condition $g(1) = 0$ to confirm our expression for $g(x)$. Finally, we calculated $g(\frac{1}{2})$ and simplified the result to match one of the given options.

Final Answer The final answer is $\boxed{{\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} \right) + {\pi \over 3}}$, which corresponds to option (A).