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

Question

2389 views around the world

You can reuse this answer
Creative Commons License

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.

Does int_1^oo1/root(3)(x+1)dxint_1^oo1/root(3)(x+1)dx converge or diverge? If it converges, what is the integral?