Question

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

Answer 1

$\frac{2}{3}(x+1)^{3/2} + \frac{2}{3}x^{3/2} + C$

Answer 2

The problem asks to evaluate the indefinite integral of the sum of two square root functions.

We need to evaluate $\int (\sqrt{x+1}+\sqrt{x})dx$.

Using the linearity property of integrals, we can split this into two separate integrals:

$$\int (\sqrt{x+1}+\sqrt{x})dx = \int \sqrt{x+1}dx + \int \sqrt{x}dx$$

Now, let's evaluate each integral:

1. For $\int \sqrt{x}dx$:
   Rewrite $\sqrt{x}$ as $x^{1/2}$.
   Using the power rule for integration, $\int u^n du = \frac{u^{n+1}}{n+1} + C$:

   $$\int x^{1/2}dx = \frac{x^{1/2+1}}{1/2+1} + C_1 = \frac{x^{3/2}}{3/2} + C_1 = \frac{2}{3}x^{3/2} + C_1$$

2. For $\int \sqrt{x+1}dx$:
   Rewrite $\sqrt{x+1}$ as $(x+1)^{1/2}$.
   This is of the form $\int (ax+b)^n dx$. For $a=1$, $b=1$, and $n=1/2$, we use the generalized power rule: $\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$.

   $$\int (x+1)^{1/2}dx = \frac{(x+1)^{1/2+1}}{1 \cdot (1/2+1)} + C_2 = \frac{(x+1)^{3/2}}{3/2} + C_2 = \frac{2}{3}(x+1)^{3/2} + C_2$$

Combining the results from both integrals:

$$\int (\sqrt{x+1}+\sqrt{x})dx = \frac{2}{3}(x+1)^{3/2} + \frac{2}{3}x^{3/2} + C$$

where $C = C_1 + C_2$ is the constant of integration.