-2\ln \left(\sqrt{\frac{\pi -1}{2}}+\sqrt{\frac{\pi +1}{2}}\right)+\sqrt{\pi -1}\sqrt{\pi +1}
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MathBot Answer:
\[-\left( 2 \cdot \ln\left( \sqrt{\frac{\pi - 1}{2}} + \sqrt{\frac{\pi + 1}{2}} \right) \right) + \sqrt{\pi - 1} \cdot \sqrt{\pi + 1} = - 2 \log{\left(\sqrt{- \frac{1}{2} + \frac{\pi}{2}} + \sqrt{\frac{1}{2} + \frac{\pi}{2}} \right)} + \sqrt{-1 + \pi} \sqrt{1 + \pi} \approx 1.16666183460850368895397726299801\]
\[\begin{aligned}-\left( 2 \cdot \ln\left( \sqrt{\frac{\pi - 1}{2}} + \sqrt{\frac{\pi + 1}{2}} \right) \right) + \sqrt{\pi - 1} \cdot \sqrt{\pi + 1}& = - 2 \cdot \log{\left(\sqrt{\left(\pi - 1\right) \cdot \frac{1}{2}} + \sqrt{\left(\pi + 1\right) \cdot \frac{1}{2}} \right)} + \sqrt{\pi - 1} \cdot \sqrt{\pi + 1}\\& = - 2 \cdot \log{\left(\sqrt{\left(-1 + \pi\right) \cdot \frac{1}{2}} + \sqrt{\left(\pi + 1\right) \cdot \frac{1}{2}} \right)} + \sqrt{\pi - 1} \cdot \sqrt{\pi + 1}\\& = - 2 \cdot \log{\left(\sqrt{- \frac{1}{2} + \frac{\pi}{2}} + \sqrt{\left(\pi + 1\right) \cdot \frac{1}{2}} \right)} + \sqrt{\pi - 1} \cdot \sqrt{\pi + 1}\\& = - 2 \cdot \log{\left(\sqrt{- \frac{1}{2} + \frac{\pi}{2}} + \sqrt{\left(1 + \pi\right) \cdot \frac{1}{2}} \right)} + \sqrt{\pi - 1} \cdot \sqrt{\pi + 1}\\& = - 2 \cdot \log{\left(\sqrt{- \frac{1}{2} + \frac{\pi}{2}} + \sqrt{\frac{1}{2} + \frac{\pi}{2}} \right)} + \sqrt{\pi - 1} \cdot \sqrt{\pi + 1}\\& = - 2 \cdot \log{\left(\sqrt{\frac{1}{2} + \frac{\pi}{2}} + \sqrt{- \frac{1}{2} + \frac{\pi}{2}} \right)} + \sqrt{\pi - 1} \cdot \sqrt{\pi + 1}\\& = -2 \cdot \log{\left(\sqrt{\frac{1}{2} + \frac{\pi}{2}} + \sqrt{- \frac{1}{2} + \frac{\pi}{2}} \right)} + \sqrt{\pi - 1} \cdot \sqrt{\pi + 1}\\& = -2 \cdot \log{\left(\sqrt{\frac{1}{2} + \frac{\pi}{2}} + \sqrt{- \frac{1}{2} + \frac{\pi}{2}} \right)} + \sqrt{-1 + \pi} \cdot \sqrt{\pi + 1}\\& = -2 \cdot \log{\left(\sqrt{\frac{1}{2} + \frac{\pi}{2}} + \sqrt{- \frac{1}{2} + \frac{\pi}{2}} \right)} + \sqrt{-1 + \pi} \cdot \sqrt{1 + \pi}\\& = -2 \cdot \log{\left(\sqrt{\frac{1}{2} + \frac{\pi}{2}} + \sqrt{- \frac{1}{2} + \frac{\pi}{2}} \right)} + \sqrt{1 + \pi} \cdot \sqrt{-1 + \pi}\end{aligned}\]