Question

高数广义积分，求解，

Answer 1

设√x=tant，原式=2∫（π/4，π/2）tant（sect）^2/[tant（sect）^2]dt
=2∫（π/4，π/2）dt=2*（π/2-π/4）=π/2

Answer 2

原式=2∫(1,+∞)1/[1+(√x)²] d√x

=2arctan√x |(1,+∞)
=2×（π/2-π/4）
=π/2