Solve this, please ?

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Answer 1 · 2018-05-16T13:55:07

The answer is #=(pi^(3/2))/(3sqrt2)=1.3125#

This is an improper integral

First calculate the indefinite integral

#I=int(sqrt(arctanx)dx)/(1+x^2)#

Let #u=arctanx#, #=>#, #du=(dx)/(1+x^2)#

Therefore,

#I=intsqrtudu=2/3u^(3/2)=2/3(arctanx)^(3/2)+C#

So, the improper integral is

#int_0^(+oo)(sqrt(arctanx)dx)/(1+x^2)#

#=lim_(p->+oo)int_0^(p)(sqrt(arctanx)dx)/(1+x^2)#

#=lim_(p->+oo)[2/3(arctanx)^(3/2)]_0^p#

#=lim_(p->+oo)[2/3(arctanp)^(3/2)-2/3(arctan0)^(3/2)]#

#=2/3(pi/2)^(3/2)#

#=(pi^(3/2))/(3sqrt2)#

#=1.3125#