Problem 22 - Essential Calculus

Question

Find the integral. $$ \int \frac{1}{x \sqrt{4 x^{2}+16}} d x $$

Step-by-Step Solution

Answer

The integral \( \int \frac{1}{x \sqrt{4 x^{2}+16}} dx \) is \( \frac{1}{4}\sqrt{4x^2+16} + C \)

Step 1 - Variable Substitution

Start by setting \( u = 4x^2 + 16 \). Differentiate \( u \) with respect to \( x \) to get \( du = 8xdx \). Solving for \( dx \), we get \( dx = \frac{du}{8x} \).

Step 2 - Rewrite the Integral

Substitute \( u \) and \( dx \) back into the integral, so it now becomes \( \int \frac{1}{x \sqrt{u}}\cdot \frac{du}{8x} \)

Step 3 - Simplify the Integral

The integral now simplifies to \( \frac{1}{8}\int \frac{1}{\sqrt{u}} du \) which is easier to solve.

Step 4 - Evaluate the Integral

The integral of \( \frac{1}{\sqrt{u}} \) is \( 2\sqrt{u} \), so the equation becomes \( \frac{1}{8} \cdot 2\sqrt{u} \)

Step 5 - Substitute Back the Original Variable

Back-substitute \( u = 4x^2 + 16 \) to get the final answer. The integral equals to \( \frac{1}{4}\sqrt{4x^2+16} \)