Let $I_1 = \int_0^1 \frac{x}{x^2+1} dx$ and $I_2 = \int_1^\infty \frac{x}{x^2+1} dx$, then
Step-by-Step Solution
Key Concept: Substitution $x = 1/t$ transforms the integral into a form that can be decomposed using partial fractions
By substitution $x = \frac{1}{t}$, we get $dx = -\frac{1}{t^2}dt$. So $I_1 = \int_1^\infty \frac{\ln t}{t(t+1)}dt = \int_1^\infty \frac{\ln t}{t} - \frac{\ln t}{t+1}dt$. This evaluates to the given expression using standard integral tables and logarithmic properties.
Correct Answer: $I_1 = \{\sin \ln x\} - \ln (0,1))$