Answer: [tex]\displaystyle \frac{1}{9(1-3x)^{3}}+C[/tex]
Where C is a constant.
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Explanation:
Apply u-substitution
[tex]u = 1 - 3x\\\\\frac{du}{dx} = -3\\\\du = -3dx\\\\dx = -\frac{du}{3}\\\\[/tex]
So,
[tex]\displaystyle \int \frac{dx}{(1-3x)^4} = \int \frac{dx}{(1-3x)^4}\\\\\\\displaystyle \int \frac{dx}{(1-3x)^4} = \int (1-3x)^{-4} \ dx\\\\\\\displaystyle \int \frac{dx}{(1-3x)^4} = \int (u)^{-4} \ \left(-\frac{du}{3}\right)\\\\\\\displaystyle \int \frac{dx}{(1-3x)^4} = -\frac{1}{3}\int (u)^{-4} \ du\\\\\\[/tex]
which further leads to
[tex]\displaystyle \int \frac{dx}{(1-3x)^4} = -\frac{1}{3}\left[\frac{1}{1+(-4)}u^{-4+1}+C\right]\\\\\\\displaystyle \int \frac{dx}{(1-3x)^4} = -\frac{1}{3}\left[-\frac{1}{3}u^{-3}+C\right]\\\\\\\displaystyle \int \frac{dx}{(1-3x)^4} = -\frac{1}{3}\left[-\frac{1}{3}(1-3x)^{-3}+C\right]\\\\\\\displaystyle \int \frac{dx}{(1-3x)^4} = \frac{1}{9(1-3x)^{3}}+C\\\\\\[/tex]
Don't forget about the +C at the end.
