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Find the indefinite integral. \( \int \frac{x}{\sqrt[3]{x+10}} d x \) \( \int \frac{x}{\sqrt[3]{x+10}} d x=\square \)

Ask by Wagner Boyd. in the United States
Nov 22,2024

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\( \frac{3}{2}x\sqrt[3]{(x+10)^{2}}-\frac{9x\sqrt[3]{(x+10)^{2}}+90\sqrt[3]{(x+10)^{2}}}{10} + C \)

Solución

Calculate the integral \( \int \frac{x}{\sqrt[3]{x+10}} dx \). Evaluate the integral by following steps: - step0: Evaluate using partial integration formula: \(\int \frac{x}{\sqrt[3]{x+10}} dx\) - step1: Evaluate the power: \(\int \frac{x}{\left(x+10\right)^{\frac{1}{3}}} dx\) - step2: Evaluate the integral: \(\int x\left(x+10\right)^{-\frac{1}{3}} dx\) - step3: Prepare for integration by parts: \(\begin{align}&u=x\\&dv=\left(x+10\right)^{-\frac{1}{3}}dx\end{align}\) - step4: Calculate the derivative: \(\begin{align}&du=dx\\&dv=\left(x+10\right)^{-\frac{1}{3}}dx\end{align}\) - step5: Evaluate the integral: \(\begin{align}&du=dx\\&v=\frac{3}{2}\left(x+10\right)^{\frac{2}{3}}\end{align}\) - step6: Substitute the values into formula: \(x\times \frac{3}{2}\left(x+10\right)^{\frac{2}{3}}-\int 1\times \frac{3}{2}\left(x+10\right)^{\frac{2}{3}} dx\) - step7: Calculate: \(\frac{3}{2}x\left(x+10\right)^{\frac{2}{3}}-\int \frac{3}{2}\left(x+10\right)^{\frac{2}{3}} dx\) - step8: Evaluate the integral: \(\frac{3}{2}x\left(x+10\right)^{\frac{2}{3}}-\frac{9\left(x+10\right)^{\frac{5}{3}}}{10}\) - step9: Simplify: \(\frac{3}{2}x\sqrt[3]{\left(x+10\right)^{2}}-\frac{9\left(x+10\right)^{\frac{5}{3}}}{10}\) - step10: Simplify: \(\frac{3}{2}x\sqrt[3]{\left(x+10\right)^{2}}-\frac{9\sqrt[3]{\left(x+10\right)^{5}}}{10}\) - step11: Expand the expression: \(\frac{3}{2}x\sqrt[3]{\left(x+10\right)^{2}}-\frac{9x\sqrt[3]{\left(x+10\right)^{2}}+90\sqrt[3]{\left(x+10\right)^{2}}}{10}\) - step12: Add the constant of integral C: \(\frac{3}{2}x\sqrt[3]{\left(x+10\right)^{2}}-\frac{9x\sqrt[3]{\left(x+10\right)^{2}}+90\sqrt[3]{\left(x+10\right)^{2}}}{10} + C, C \in \mathbb{R}\) The indefinite integral of \( \int \frac{x}{\sqrt[3]{x+10}} dx \) is \( \frac{3}{2}x\sqrt[3]{(x+10)^{2}}-\frac{9x\sqrt[3]{(x+10)^{2}}+90\sqrt[3]{(x+10)^{2}}}{10} + C \), where \( C \) is an arbitrary constant.

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To solve the integral \( \int \frac{x}{\sqrt[3]{x+10}} \, dx \), we can use a substitution method. Let's set \( u = x + 10 \). Therefore, \( du = dx \) and \( x = u - 10 \). Our integral becomes: \[ \int \frac{u - 10}{\sqrt[3]{u}} \, du \] This can be separated into two simpler integrals: \[ \int (u^{2/3} - 10u^{-1/3}) \, du \] Next, we integrate each term: \[ \int u^{2/3} \, du = \frac{u^{5/3}}{5/3} = \frac{3}{5} u^{5/3} \] \[ \int -10u^{-1/3} \, du = -10 \cdot \frac{u^{2/3}}{2/3} = -15 u^{2/3} \] Putting this all together: \[ \int \frac{x}{\sqrt[3]{x+10}} \, dx = \frac{3}{5} u^{5/3} - 15 u^{2/3} + C \] Now substitute back \( u = x + 10 \): \[ = \frac{3}{5} (x + 10)^{5/3} - 15 (x + 10)^{2/3} + C \] So the final answer is: \[ \int \frac{x}{\sqrt[3]{x+10}} \, dx = \frac{3}{5} (x + 10)^{5/3} - 15 (x + 10)^{2/3} + C \] Dive into the fascinating world of integrals! They have been studied for centuries, transforming the way we calculate areas under curves. The concept of integration dates back to ancient Greece, but it didn't find its formal definition until the work of Newton and Leibniz in the 17th century. In real-world applications, integrals help in various fields such as physics, economics, and engineering. They are crucial for calculating things like the area beneath a curve (think of plotting profits over time), finding volumes of solids, and even in determining probabilities in statistics. Integrals are not just academic exercises; they play a significant role in day-to-day problem-solving!

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