Q. 26 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.) 

$\int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\frac{3}{4}(x^{4/3}+2x^{2/3})+C$.

Step 1. Given Information

Solving the given integrals. 

$\int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x$

Step 2. Solving the given integral using algebra.

$$\begin{gathered} \int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\int \left(\frac{x^{2/3}}{x^{1/3}}+\frac{1}{x^{1/3}}\right)\mathrm{d}x \\ \int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\int \left(x^{2/3}·x^{-1/3}+x^{-1/3}\right)\mathrm{d}x \\ \int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\int \left(x^{2/3-1/3}+x^{-1/3}\right)\mathrm{d}x \\ \int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\int \left(x^{1/3}+x^{-1/3}\right)\mathrm{d}x \end{gathered}$$

Step 3. After simplifying

$$\begin{gathered} \int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\int x^{1/3}\mathrm{d}x+\int x^{-1/3}\mathrm{d}x \\ \int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\left[\frac{x^{1/3+1}}{1/3+1}\right]+\left[\frac{x^{-1/3+1}}{-1/3+1}\right]+C \\ \int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\left[\frac{x^{4/3}}{4/3}\right]+\left[\frac{x^{2/3}}{2/3}\right]+C \\ \int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\frac{3}{4}·x^{4/3}+\frac{3}{2}·x^{2/3}+C \\ \int \frac{x^{2/3}+1}{\sqrt[3]{x}}\mathrm{d}x=\frac{3}{4}(x^{4/3}+2x^{2/3})+C \end{gathered}$$