Question

integral 0 to 1 log x/(1+x^2) dx

Answer 1

Answer:

logx√1−x2dx =logx∫1√1−x2dx−∫1x(∫1√1−x2dx)dx Now, ∫1√1−x2dx=arcsinx+c But inputting this into the ∫1x∫(1√1 ...