Integration using Substitution
To solve the integral
\[
\int \frac{dx}{1 + \sqrt{x}},
\]
we can use a substitution. Let's set \( u = \sqrt{x} \), then \( u^2 = x \). Now, we differentiate both sides with respect to \( x \) to get the differential relationship between \( u \) and \( x \). This gives us \( 2u du = dx \), or \( dx = 2u du \).
Substitute into the integral:
\[
\int \frac{2u}{1 + u} du.
\]
Now, let's divide the numerator by the denominator to simplify:
\[
\int \frac{2u}{1 + u} du = \int \left(2 - \frac{2}{1 + u}\right) du.
\]
Breaking this integral apart we get:
\[
\int 2 du - \int \frac{2}{1 + u} du.
\]
Integrating term by term:
\[
2u - 2\ln|1 + u| + C.
\]
Now we substitute back to express in terms of \( x \):
\[
2\sqrt{x} - 2\ln|\sqrt{x} + 1| + C,
\]
where \( C \) is the constant of integration.
Therefore, the solution to the integral is:
\[
\int \frac{dx}{1 + \sqrt{x}} = 2\sqrt{x} - 2\ln|\sqrt{x} + 1| + C.
\]
