Using the integral test, how do you show whether #n/(n^2+1)# diverges or converges? Calculus Tests of Convergence / Divergence Integral Test for Convergence of an Infinite Series 1 Answer Jim H May 1, 2015 #int_1^oo x/(x^2+1) dx# Let #u=x^2+1#, so that #du= 2x dx# #int x/(x^2+1) dx = 1/2 int (2x)/(x^2+1) dx = 1/2 1/u du = 1/2 ln abs(u) +C# #int_1^oo x/(x^2+1) dx =lim_(brarr oo) 1/2ln(x^2+1)|_1^b # which diverges, So the series diverges. Answer link Related questions What is the Integral Test for Convergence of an Infinite Series? How do you know when to use the integral test for an infinite series? How do you use the Integral test on the infinite series #sum_(n=1)^oo1/root5(n)# ? How do you use the Integral test on the infinite series #sum_(n=1)^oo1/n^5# ? How do you use the Integral test on the infinite series #sum_(n=1)^oo1/(2n+1)^3# ? How do you use the Integral test on the infinite series #sum_(n=1)^oo1/sqrt(n+4)# ? How do you determine if the series #ln(1/2) + ln(1/3) + ln(3/4) + ... +ln[k/(k + 1)] + ....# converges? How do you know #{-1,1,-1,1,-1,1,...}# converges or diverges? Using the integral test, how do you show whether # (1 + (1/x))^x# diverges or converges? Using the integral test, how do you show whether #sum 1/(n^2+1)# diverges or converges from n=1... See all questions in Integral Test for Convergence of an Infinite Series Impact of this question 6675 views around the world You can reuse this answer Creative Commons License