How do you find the integral of #int x/((x^2 +2)^2) dx# from 0 to infinity? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Tom Oct 26, 2015 #int_0^oo x/(x^2+2)^2dx = 1/2int_0^oo(2x)/(x^2+2)^2dx# Let's #u = x^2+2# So #du = 2xdx# #1/2int_2^oo1/(u^2)du# #-1/2[1/u]_2^oo# #=-1/2(1/oo-1/2) = -1/2(0-1/2)# #=1/4# 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 9962 views around the world You can reuse this answer Creative Commons License