Question

use a change of variables to find \\(\int \frac{1}{6x - 1} \\, dx\\)

Explanation:

Step1: Choose substitution

Let \( u = 6x - 1 \). Then, find the derivative of \( u \) with respect to \( x \).
The derivative \( \frac{du}{dx}=6 \), so \( dx=\frac{du}{6} \).

Step2: Substitute into integral

Substitute \( u = 6x - 1 \) and \( dx=\frac{du}{6} \) into the integral \( \int\frac{1}{6x - 1}dx \).
We get \( \int\frac{1}{u}\cdot\frac{du}{6} \).

Step3: Simplify and integrate

Simplify the integral: \( \frac{1}{6}\int\frac{1}{u}du \).
The integral of \( \frac{1}{u} \) with respect to \( u \) is \( \ln|u|+C \) (where \( C \) is the constant of integration).
So, \( \frac{1}{6}\int\frac{1}{u}du=\frac{1}{6}\ln|u| + C \).

Step4: Substitute back

Substitute \( u = 6x - 1 \) back into the expression.
We get \( \frac{1}{6}\ln|6x - 1|+C \).

Answer:

\(\frac{1}{6}\ln|6x - 1| + C\) (where \( C \) is the constant of integration)