Does $\int_{1}^{\infty} \frac{1}{\sqrt[3]{x + 1}} d x$ $int_1^oo1/root(3)(x+1)dx$ converge or diverge? If it converges, what is the integral?

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Does $\int_{1}^{\infty} \frac{1}{\sqrt[3]{x + 1}} d x$ $int_1^oo1/root(3)(x+1)dx$ converge or diverge? If it converges, what is the integral?

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Answer 1 · 2017-02-06T06:22:17

The integral diverges.

Explanation:

Let $u = x + 1 \Rightarrow d u = d x$ $u = x+1 => du = dx$ . At $x = 1$ $x=1$ , we have $u = 2$ $u = 2$ . As $x \to \infty$ $x->oo$ , we have $u \to \infty$ $u->oo$ . Using this substitution,

$\int_{1}^{\infty} \frac{1}{\sqrt[3]{x + 1}} d x = \int_{1}^{\infty} {(x + 1)}^{- \frac{1}{3}} d x$ $int_1^oo1/root(3)(x+1)dx = int_1^oo(x+1)^(-1/3)dx$

$= \int_{2}^{\infty} u^{- \frac{1}{3}} d u$ $=int_2^oou^(-1/3)du$

$= {[\frac{u^{\frac{2}{3}}}{\frac{2}{3}}]}_{2}^{\infty}$ $=[u^(2/3)/(2/3)]_2^oo$

$= \frac{3}{2} (\infty^{\frac{2}{3}} - 2^{\frac{2}{3}})$ $=3/2(oo^(2/3)-2^(2/3))$

$= \infty$ $=oo$

Thus the integral diverges.

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Does $\int_{1}^{\infty} \frac{1}{\sqrt[3]{x + 1}} d x$ $int_1^oo1/root(3)(x+1)dx$ converge or diverge? If it converges, what is the integral?

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Does $\int_{1}^{\infty} \frac{1}{\sqrt[3]{x + 1}} d x$ $int_1^oo1/root(3)(x+1)dx$ converge or diverge? If it converges, what is the integral?