/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Find the integral. $$ \int \... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 22

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

Short Answer

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

Step by step solution

01

- Variable Substitution

Start by setting \( u = 4x^2 + 16 \). Differentiate \( u \) with respect to \( x \) to get \( du = 8xdx \). Solving for \( dx \), we get \( dx = \frac{du}{8x} \).
02

- Rewrite the Integral

Substitute \( u \) and \( dx \) back into the integral, so it now becomes \( \int \frac{1}{x \sqrt{u}}\cdot \frac{du}{8x} \)
03

- Simplify the Integral

The integral now simplifies to \( \frac{1}{8}\int \frac{1}{\sqrt{u}} du \) which is easier to solve.
04

- Evaluate the Integral

The integral of \( \frac{1}{\sqrt{u}} \) is \( 2\sqrt{u} \), so the equation becomes \( \frac{1}{8} \cdot 2\sqrt{u} \)
05

- Substitute Back the Original Variable

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

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