Question

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{1+\sqrt{x}}{x} d x$$

Short answer

Question: Determine the indefinite integral of the function \(f(x) = \frac{1+\sqrt{x}}{x}\). Answer: The indefinite integral of the function \(f(x) = \frac{1+\sqrt{x}}{x}\) is \(\int f(x) dx = \ln{|x|} + 2x^{\frac{1}{2}} + C\), where C is an arbitrary constant.

Step-by-step solution

Rewrite the integrand

First, let's rewrite the integrand by splitting it into two fractions: $$\int \frac{1+\sqrt{x}}{x} d x = \int \frac{1}{x} d x + \int \frac{\sqrt{x}}{x} d x$$ Now simplify the second fraction: \(\frac{\sqrt{x}}{x} = \frac{x^{\frac{1}{2}}}{x^1} = x^{-\frac{1}{2}}\). So, our integrand becomes: $$\int \frac{1}{x} d x + \int x^{-\frac{1}{2}} d x$$

Integrate each term separately

Now we can integrate each term using basic integration rules: 1. For the first term: \(\int \frac{1}{x} dx = \ln{|x|} + C_1\) 2. For the second term: \(\int x^{-\frac{1}{2}} dx = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C_2 = 2x^{\frac{1}{2}} + C_2\)

Combine the two terms

Now we need to combine the two terms that we've integrated: $$\int \frac{1+\sqrt{x}}{x} d x = \ln{|x|} + 2x^{\frac{1}{2}} + C$$ Where \(C = C_1 + C_2\) is an arbitrary constant.

Check the work by differentiation

Now we need to differentiate our result to verify that it matches the original integrand. Differentiating the result with respect to \(x\) gives us: $$\frac{d}{d x} (\ln{|x|} + 2x^{\frac{1}{2}} + C) = \frac{1}{x} + 2 \cdot \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{x} + x^{-\frac{1}{2}}$$ And combining the terms as in the original integrand, we get: $$\frac{1+\sqrt{x}}{x} = \frac{1}{x} + x^{-\frac{1}{2}}$$ As our differentiation matches the original integrand, we can confirm that our solution is correct: $$\int \frac{1+\sqrt{x}}{x} d x = \ln{|x|} + 2x^{\frac{1}{2}} + C$$