To compute the integral

$\int \frac{1}{\sqrt{x} + 2x} \, dx,$

we'll proceed step by step:

Step 1: Substitution

Let $u = \sqrt{x}$, which implies: $x = u^2 \quad \text{and} \quad dx = 2u \, du.$

Substitute these into the integral: $\int \frac{1}{\sqrt{x} + 2x} \, dx = \int \frac{1}{u + 2u^2} \cdot 2u \, du.$

Simplify the expression: $\int \frac{2u}{u + 2u^2} \, du = \int \frac{2}{1 + 2u} \, du.$

Step 2: Simplify the integral

This reduces to: $\int \frac{2}{1 + 2u} \, du.$

Let $v = 1 + 2u$, so $dv = 2 \, du$. Substituting, we have: $\int \frac{2}{1 + 2u} \, du = \int \frac{1}{v} \, dv.$

Step 3: Solve the integral

The integral of $\frac{1}{v}$ is: $\ln|v| + C,$ where $C$ is the constant of integration. Substituting back $v = 1 + 2u$: $\ln|1 + 2u| + C.$

Step 4: Back-substitute $u = \sqrt{x}$

Finally, replace $u$ with $\sqrt{x}$ to return to the original variable $x$: $\ln|1 + 2\sqrt{x}| + C.$

Final Answer:

$\int \frac{1}{\sqrt{x} + 2x} \, dx = \ln|1 + 2\sqrt{x}| + C.$

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Would you like further details or clarification on this computation?

Related Questions:

1. What if the denominator had a higher power of $x$?
2. How do substitutions help simplify integrals like this one?
3. Can you explain why logarithms appear in integrals of this form?
4. What are the limitations of substitution methods?
5. How would you integrate $\frac{1}{\sqrt{x} + 3x}$ instead?

Tip:

When dealing with integrals involving square roots, substitutions like $u = \sqrt{x}$ are often useful for simplifying the expression.