What is #int_1^oo 1/((1+x)(x^(1/2)))dx#?

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Answer 1 · 2015-11-09T20:14:48

#int_1^oo 1/((1+x)sqrt(x))dx = 2int_1^oo 1/(2(1+x)sqrt(x))#

#u = sqrt(x)#
#u^2=x#

if #sqrt(x) = oo# then #u = oo#
if #sqrt(x) = 1# then #u = 1#

#du = 1/(2sqrt(x))dx#

So now we have

#int_1^oo 1/(1+u^2)du#

#= 2[arctan(u)]_1^oo#

#= 2(arctan(oo) - arctan(1))#

#=2(pi/2-pi/4) = pi/2#