Question

use the substitution ( u = 8x^{2}+x ) to evaluate the following indefinite integral.
int(16 x + 1) sqrt{8 x^{2}+x} d x
write the integrand in terms of ( u ).
int(16 x + 1) sqrt{8 x^{2}+x} d x=int(square) d u
evaluate the integral.
int(16 x + 1) sqrt{8 x^{2}+x} d x=square
(type an expression using ( x ) as the variable.)

Explanation:

Step1: Find \( du \)

Given \( u = 8x^2 + x \), differentiate both sides with respect to \( x \).
The derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = \frac{d}{dx}(8x^2 + x) \).
Using the power rule, \( \frac{d}{dx}(8x^2) = 16x \) and \( \frac{d}{dx}(x) = 1 \), so \( \frac{du}{dx} = 16x + 1 \).
Then, \( du = (16x + 1)dx \).

Step2: Rewrite the integral in terms of \( u \)

The original integral is \( \int (16x + 1)\sqrt{8x^2 + x} \, dx \).
Since \( u = 8x^2 + x \), then \( \sqrt{8x^2 + x} = \sqrt{u} = u^{\frac{1}{2}} \), and from Step 1, \( (16x + 1)dx = du \).
So the integral becomes \( \int u^{\frac{1}{2}} \, du \).

Step3: Evaluate the integral with respect to \( u \)

Integrate \( u^{\frac{1}{2}} \) with respect to \( u \) using the power rule for integration: \( \int u^n \, du = \frac{u^{n + 1}}{n + 1} + C \) (where \( n
eq -1 \)).
Here, \( n = \frac{1}{2} \), so \( \int u^{\frac{1}{2}} \, du = \frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C = \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{2}{3}u^{\frac{3}{2}} + C \).

Step4: Substitute back \( u = 8x^2 + x \)

Replace \( u \) with \( 8x^2 + x \) in the result from Step 3.
We get \( \frac{2}{3}(8x^2 + x)^{\frac{3}{2}} + C \).

Answer:

For the first blank (rewriting the integral in terms of \( u \)): \( \boldsymbol{u^{\frac{1}{2}}} \) (or \( \boldsymbol{\sqrt{u}} \))

For the integral evaluation: \( \boldsymbol{\frac{2}{3}(8x^2 + x)^{\frac{3}{2}} + C} \) (where \( C \) is the constant of integration)