What is #int_1^oo 1/((1+x)(x^(1/2)))dx#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Tom Nov 9, 2015 #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# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1429 views around the world You can reuse this answer Creative Commons License