Question - Integration using Substitution
Solution:
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:$$\int \frac{dx}{1 + \sqrt{x}} = 2\sqrt{x} - 2\ln|\sqrt{x} + 1| + C.$$
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