Does #int_1^oo1/root(3)(x+1)dx# converge or diverge? If it converges, what is the integral?

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Answer 1 · 2017-02-06T06:22:17

The integral diverges.

Let #u = x+1 => du = dx#. At #x=1#, we have #u = 2#. As #x->oo#, we have #u->oo#. Using this substitution,

#int_1^oo1/root(3)(x+1)dx = int_1^oo(x+1)^(-1/3)dx#

#=int_2^oou^(-1/3)du#

#=[u^(2/3)/(2/3)]_2^oo#

#=3/2(oo^(2/3)-2^(2/3))#

#=oo#

Thus the integral diverges.