consider the indefinite integral $\\int \\frac{x - 3}{(3x - 2)^2}dx$. the substitution $u = 3x - 2$…

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

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

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)