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If $\mathrm{I}=\int \frac{2 x-7}{\sqrt{3 x-2}} \mathrm{~d} x$, then $\mathrm{I}$ is given by
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The correct answer is:
$\frac{4}{27}(3 x-2)^{\frac{3}{2}}-\frac{34}{9}(3 x-2)^{\frac{1}{2}}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
$\begin{aligned} \mathrm{I} & =\int \frac{2 x-7}{\sqrt{3 x-2}} \mathrm{~d} x \\ & =\int \frac{\frac{2}{3}(3 x-2)-\frac{17}{3}}{\sqrt{3 x-2}} \mathrm{~d} x \\ & =\frac{2}{3} \int(3 x-2)^{\frac{1}{2}} \mathrm{~d} x-\frac{17}{3} \int(3 x-2)^{\frac{-1}{2}} \mathrm{~d} x \\ & =\frac{2}{3} \times \frac{(3 x-2)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{3}-\frac{17}{3} \times \frac{(3 x-2)^{\frac{1}{2}}}{\frac{1}{2}} \times \frac{1}{3}+\mathrm{c} \\ & =\frac{4}{27}(3 x-2)^{\frac{3}{2}}-\frac{34}{9}(3 x-2)^{\frac{1}{2}}+\mathrm{c}\end{aligned}$
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