consider the indefinite integral $\int \frac{x - 3}{(3x - 2)^2}dx$. the substitution $u = 3x - 2$ transforms the integral into:
$\bigcirc \int \frac{1}{u^2}du$
$\bigcirc \int \frac{\frac{u - 2}{3} - 3}{u^2}du$
$\bigcirc$ none of these options are correct.
$\bigcirc \int \frac{u - \frac{7}{3}}{u^2}du$
$\bigcirc \int \frac{u - 7}{9u^2}du$

Question

Accepted Answer

# Explanation:
## Step1: Find $dx$ in terms of $du$
Given $u = 3x-2$, then $du=3dx$, so $dx=\frac{1}{3}du$. Also, $x=\frac{u + 2}{3}$.
## Step2: Substitute $x$ and $dx$ into the integral
Substitute $x=\frac{u + 2}{3}$ and $dx=\frac{1}{3}du$ into $\int\frac{x - 3}{(3x - 2)^2}dx$.
$$
\begin{align*}
\int\frac{x - 3}{(3x - 2)^2}dx&=\int\frac{\frac{u + 2}{3}-3}{u^2}\cdot\frac{1}{3}du\
&=\int\frac{u + 2-9}{9u^2}du\
&=\int\frac{u-7}{9u^2}du
\end{align*}
$$

# Answer:
$\int\frac{u - 7}{9u^2}du$ (the fifth option)

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QUESTION IMAGE

Question

Explanation:

Step1: Find \(dx\) in terms of \(du\)

Step2: Substitute \(x\) and \(dx\) into the integral

Answer: