Problems

Solution: First suppose \(|a| < 1\), then \begin{align*} && I &= \int_0^1 \frac{1}{x^2+2ax+1} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2 +1-a^2} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2 +(\sqrt{1-a^2})^2} \d x \tag{\(1-a^2 > 0\)}\\ &&&= \left [\frac{1}{\sqrt{1-a^2}} \tan^{-1} \frac{x+a}{\sqrt{1-a^2}} \right]_0^1 \\ &&&= \frac{1}{\sqrt{1-a^2}} \left ( \tan^{-1} \frac{a+1}{\sqrt{1-a^2}} - \tan^{-1} \frac{a}{\sqrt{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{\frac{a+1}{\sqrt{1-a^2}}-\frac{a}{\sqrt{1-a^2}}}{1+\frac{(a+1)a}{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{\frac{1}{\sqrt{1-a^2}}}{\frac{1+a}{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{1-a}{\sqrt{1-a^2}}\right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \sqrt { \frac{1-a}{1+a}} \\ \end{align*} Second, suppose \(|a| > 1\), then \begin{align*} && I &= \int_0^1 \frac{1}{x^2+2ax+1} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2-(a^2-1)} \d x \\ &&&= \int_0^1 \frac{1}{(x+a-\sqrt{a^2-1})(x+a+\sqrt{a^2-1})} \d x \tag{\(a^2-1 > 0\)} \\ &&&= \frac{1}{2\sqrt{a^2-1}}\int_0^1 \left ( \frac{1}{x+a-\sqrt{a^2-1}} - \frac{1}{x+a+\sqrt{a^2-1}} \right) \d x \\ &&&= \frac{1}{2\sqrt{a^2-1}} \left [ \ln |x+a-\sqrt{a^2-1}|- \ln |x+a+\sqrt{a^2-1}| \right]_0^1 \\ &&&= \frac{1}{2\sqrt{a^2-1}} \left ( \ln |1+a-\sqrt{a^2-1}| - \ln|1+a+\sqrt{a^2-1}| - \ln|a-\sqrt{a^2-1}| +\ln|a + \sqrt{a^2-1}| \right) \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln | \frac{(1+a-\sqrt{a^2-1})(a+\sqrt{a^2-1})}{(1+a+\sqrt{a^2-1})(a-\sqrt{a^2-1})}|\\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{a+a^2-(a^2-1) +\sqrt{a^2-1}}{1+a-\sqrt{a^2-1}}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{(1+a +\sqrt{a^2-1})^2}{(1+a)^2-(a^2-1)}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{1+2a+a^2+a^2-1+2(1+a)\sqrt{a^2-1}}{2+2a}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |a+\sqrt{a^2-1}| \\ \end{align*}