Find out the answer to "\int_0^1 (1)/(1+\root(3)(x))dx"

Answered question

2022-03-17

\int_0^1 (1)/(1+\root(3)(x))dx

Answer & Explanation

Vasquez

Expert2023-04-24Added 669 answers

To solve the integral, we can use the substitution method. Let's substitute $u = 1 + \sqrt[3]{x}$ , which means that $\frac{d u}{d x} = (\frac{1}{3}) x^{- \frac{2}{3}}$ , or $d x = 3 u^{2} {(\sqrt[3]{u} - 1)}^{2} d u$ .

Now, we can rewrite the integral in terms of u:

$\int [0, 1] (\frac{1}{1 + \sqrt[3]{x}}) d x$
$= \int [1, 2] (\frac{1}{1 + u}) 3 u^{2} {(\sqrt[3]{u} - 1)}^{2} d u$

The bounds of integration have changed because when $x = 0, u = 1 + \sqrt[3]{0} = 1$ , and when $x = 1, u = 1 + \sqrt[3]{1} = 2$ .

Simplifying the integrand, we get:
$= 3 \int [1, 2] \frac{u^{\frac{2}{3}} {(\sqrt[3]{u} - 1)}^{2}}{1 + u} d u$

To proceed, let's use partial fractions to break up the integrand:
$u^{\frac{2}{3}} \frac{{(\sqrt[3]{u} - 1)}^{2}}{1 + u}$
$= \frac{A}{1 + u} + B \frac{u^{\frac{1}{3}}}{1 + u} + C \frac{u^{\frac{2}{3}}}{{(1 + u)}^{2}}$

Solving for A, B, and C, we get:
$A = \frac{1}{3}$
$B = - \frac{2}{3}$
$C = \frac{2}{3}$

Therefore, we can rewrite the integral as:
$= 3 \int [1, 2] [\frac{\frac{1}{3}}{1 + u} - (\frac{2}{3}) \frac{u^{\frac{1}{3}}}{1 + u} + (\frac{2}{3}) \frac{u^{\frac{2}{3}}}{{(1 + u)}^{2}}] d u$

Now, we can integrate each term separately:
$= 3 [(\frac{1}{3}) \ln | 1 + u | - (\frac{4}{9}) \frac{u^{\frac{4}{3}}}{1 + u} + (\frac{2}{3}) \ln | 1 + u | - (\frac{4}{3}) (\frac{1}{1 + u})] ∣_{1}^{2}$

Plugging in the limits of integration and simplifying, we get:
$= 3 [\ln (2) - \ln (\frac{2}{∛} 2) + \frac{2}{9} (2^{\frac{4}{3}} - 1) + \frac{8}{9} \ln (2) - \frac{8}{9} \ln (3)]$
$= 3 [\ln (\frac{2}{\sqrt[3]{2}}) + \frac{2}{9} (2^{\frac{4}{3}} - 1) + \frac{8}{9} \ln (3)]$

Thus, the value of the integral is approximately 1.46.

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