Q. 34 - 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.) 

$\int \frac{\sqrt{x}+1}{x}\mathrm{d}x$

Step-by-Step Solution

Answer

The solution of the given integral is $\int \frac{\sqrt{x}+1}{x}\mathrm{d}x=2\sqrt{x}+\ln x+C$.

Step 1. Given Information

Solving the given integrals. 

$\int \frac{\sqrt{x}+1}{x}\mathrm{d}x$

Step 2. Solving the given integral using algebra.

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

Step 3. After simplifying

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