Question

定积分计算题求解

Answer 1

见下图：

Answer 2

用换元法来解，令x=tgt/2，dx=(sect)^2dt/2，0<=x<=1，则0<=t<=arctg2。
原积分=∫(0->arctg2)sectdt/2
=∫(0->arctg2)dt/2cost
=∫(0->arctg2)costdt/2(cost)^2
=∫(0->arctg2)d(sint)/2(1-(sint)^2)
=∫(0->arctg2)(1/(1+sint)+1/(1-sint))d(sint)/4
=(0->arctg2)ln[(1+sint)/(1-sint)]/4
=ln(√5+2)/2

Answer 3

let
x=(1/2)tanu
dx=(1/2)(secu)^2 du
x=0, u=0
x=1, u=arctan(2)
∫(0->1) 1/√(1+4x^2) dx
=∫(0->arctan(2)) (1/2)(secu)^2 du / (secu)
=(1/2)∫(0->arctan(2)) secu du
=(1/2)[ln|secu +tanu|]|(0->arctan(2))
=(1/2) ln(√5 + 2)