How do I find the indefinite integral: $\displaystyle \int\frac{5}{{(3x-1)}^3} dx $
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Originally Posted by millerst How do I find the indefinite integral: $\displaystyle \int\frac{5}{(3x-1)^3} dx $ I've aded the brackets that should have been included. Substitute u = 3x - 1.
make $\displaystyle u = 3x-1 \Rightarrow \frac{du}{dx} = 3$ therefore $\displaystyle \frac{5}{3}\int \frac{du}{dx}\frac{1}{u^3}~dx$ Can you take it from here?
$\displaystyle \int\frac{5}{(3x-1)^3} dx $ So the answer works out to? $\displaystyle \frac{-5}{6(3x-1)^2} + C $
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