Q. 27 - Calculus

Question

Use integration formulas to solve each integral in Exercises 21–62. You may have to use algebra, educated guess and-check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating. 

$\int \frac{x+1}{\sqrt{x}}dx$

Step-by-Step Solution

Answer

The answer is $\frac{2x^{\frac{3}{2}}}{3}+2x^{\frac{1}{2}}+C.$

Step 1. Given Information.

The given integral is $\int \frac{x+1}{\sqrt{x}}dx$.

Step 2. Integration.

On integrating, we get,

$$\begin{gathered} \int \left(\frac{x+1}{\sqrt{x}}\right)dx \\ =\int \left(\frac{x}{\sqrt{x}}+\frac{1}{\sqrt{x}}\right)dx \\ =\int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx \\ =\int \sqrt{x}dx+\int \frac{1}{\sqrt{x}}dx \\ =\int x^{\frac{1}{2}}dx+\int x^{-\frac{1}{2}}dx \\ =\left(\frac{2x^{\frac{3}{2}}}{3}\right)+2x^{\frac{1}{2}}+C \end{gathered}$$

Step 3. Verification.

On Differentiating $\left(\frac{2x^{\frac{3}{2}}}{3}\right)+2x^{\frac{1}{2}}+C,$ we get,

$$\begin{gathered} \frac{3x^{{\displaystyle \frac{1}{2}}}}{3}+x^{-\frac{1}{2}} \\ =\sqrt{x}+\frac{1}{\sqrt{x}} \\ =\frac{x+1}{\sqrt{x}} \end{gathered}$$

Hence Proved.