The worst acquisition in history, again

### --- ## 3. Step-by-step solution Let's solve the integral step by step. ### Step 1: The function to integrate \[ I = \int_{\sqrt{3}}^{\sqrt{8}} \frac{x}{x^2 + 1} \, dx} \] ### Step 2: Use substitution Let \( u = x^2 + 1 \). Then, \( du = 2x \, dx \) → \( x \, dx = \frac{du}{2} \) ### Step 3: Change the limits - When \( x = \sqrt{3} \): \( u = (\sqrt{3})^2 + 1 = 3 + 1 = 4 \) - When \( x = \sqrt{8} \): \( u = (\sqrt{8})^2 + 1 = 8 + 1 = 9 \) ### Step 4: Rewrite the integral \[ I = \int_{u=4}^{u=9}} \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int_{4}^{9} \frac{1}{u} \, du} \] ### Step 5: Evaluate the integral \[ \frac{1}{2} \left[ \ln|u| \right]_{4}^{9} = \frac{1}{2} (\ln 9 - \ln 4) = \frac{1}{2} \ln\left(\frac{9}{4}\right) \] ### Step 6: Simplify the result \[ \frac{1}{2} \ln\left(\frac{9}{4}\right) = \frac{1}{2} (\ln 9 - \ln 4) \] We can write this as: \[ \frac{1}{2} (\ln(3^2) - \ln(2^2)) = \frac{1}{2} (2\ln 3 - 2\ln 2) = \ln 3 - \ln 2 = \ln\left(\frac{3}{2}\right) \] So, the area is \( \ln\left(\frac{3}{2}\right) \). --- ## 4. What does this mean? The result is approximately 0.405