## Problem Statement

Calculate the definite integral:

$$
\int_{2}^{9} \frac{x}{\sqrt[3]{x-1}} d x
$$
## Solution

$$
\begin{aligned}
& \int_{2}^{9} \frac{x}{\sqrt[3]{x-1}} d x=\int_{2}^{9} \frac{x-1+1}{\sqrt[3]{x-1}} d x=\int_{2}^{9} \sqrt[3]{(x-1)^{2}} d x+\int_{2}^{9} \frac{1}{\sqrt[3]{x-1}} d x= \\
& =\left.\frac{3}{5} \cdot \sqrt[3]{(x-1)^{5}}\right|_{2} ^{9}+\left.\frac{3}{2} \cdot \sqrt[3]{(x-1)^{2}}\right|_{2} ^{9}= \\
& =\frac{3}{5} \cdot \sqrt[3]{(9-1)^{5}}-\frac{3}{5} \cdot \sqrt[3]{(2-1)^{5}}+\frac{3}{2} \cdot \sqrt[3]{(9-1)^{2}}-\frac{3}{2} \cdot \sqrt[3]{(2-1)^{2}}= \\
& =\frac{3}{5} \cdot \sqrt[3]{8^{5}}-\frac{3}{5} \cdot \sqrt[3]{1^{5}}+\frac{3}{2} \cdot \sqrt[3]{8^{2}}-\frac{3}{2} \cdot \sqrt[3]{1^{2}}= \\
& =\frac{3}{5} \cdot 2^{5}-\frac{3}{5}+\frac{3}{2} \cdot 2^{2}-\frac{3}{2}=\frac{96}{5}-\frac{3}{5}+6-\frac{3}{2}= \\
& =\frac{93}{5}+\frac{9}{2}=\frac{186}{10}+\frac{45}{10}=\frac{231}{10}=23.1
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_4-31"

Categories: Kuznetsov's Integral Problems Problem 4 | Integrals

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