Q. 50 - 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.)

$\int (x^2+1)\sqrt{x}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int (x^2+1)\sqrt{x}\mathrm{d}x=\frac{2}{7}{x}^{7/2}+\frac{2}{3}{x}^{3/2}+C$.

Step 1. Given Information

Solving the given integrals. 

$\int (x^2+1)\sqrt{x}\mathrm{d}x$

Step 2. Simplifying the given integral.

$$\begin{gathered} \int (x^2+1)\sqrt{x}\mathrm{d}x=\int (x^2+1)x^{1/2}\mathrm{d}x \\ \int (x^2+1)\sqrt{x}\mathrm{d}x=\int (x^2·x^{1/2}+1·x^{1/2})\mathrm{d}x \\ \int (x^2+1)\sqrt{x}\mathrm{d}x=\int (x^{2+1/2}+x^{1/2})\mathrm{d}x \\ \int (x^2+1)\sqrt{x}\mathrm{d}x=\int (x^{5/2}+x^{1/2})\mathrm{d}x \end{gathered}$$

Step 3. After simplification.

$$\begin{gathered} \int (x^2+1)\sqrt{x}\mathrm{d}x=\int x^{5/2}\mathrm{d}x+\int x^{1/2}\mathrm{d}x \\ \int (x^2+1)\sqrt{x}\mathrm{d}x=\left[\frac{x^{5/2+1}}{5/2+1}\right]+\left[\frac{x^{1/2+1}}{1/2+1}\right]+C \\ \int (x^2+1)\sqrt{x}\mathrm{d}x=\left[\frac{x^{7/2}}{7/2}\right]+\left[\frac{x^{3/2}}{3/2}\right]+C \\ \int (x^2+1)\sqrt{x}\mathrm{d}x=\frac{2}{7}{x}^{7/2}+\frac{2}{3}{x}^{3/2}+C \end{gathered}$$