Question

Another question from me ahheheheheheh help pls

Answer 1

Answer:
`\displaystyle (1)/(9(1-3x)^(3))+C`

Where C is a constant.

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Step-by-step explanation:

Apply u-substitution

`u = 1 - 3x\\\\(du)/(dx) = -3\\\\du = -3dx\\\\dx = -(du)/(3)\\\\`

So,

`\displaystyle \int (dx)/((1-3x)^4) = \int (dx)/((1-3x)^4)\\\\\\\displaystyle \int (dx)/((1-3x)^4) = \int (1-3x)^(-4) \ dx\\\\\\\displaystyle \int (dx)/((1-3x)^4) = \int (u)^(-4) \ \left(-(du)/(3)\right)\\\\\\\displaystyle \int (dx)/((1-3x)^4) = -(1)/(3)\int (u)^(-4) \ du\\\\\\`

which further leads to

`\displaystyle \int (dx)/((1-3x)^4) = -(1)/(3)\left[(1)/(1+(-4))u^(-4+1)+C\right]\\\\\\\displaystyle \int (dx)/((1-3x)^4) = -(1)/(3)\left[-(1)/(3)u^(-3)+C\right]\\\\\\\displaystyle \int (dx)/((1-3x)^4) = -(1)/(3)\left[-(1)/(3)(1-3x)^(-3)+C\right]\\\\\\\displaystyle \int (dx)/((1-3x)^4) = (1)/(9(1-3x)^(3))+C\\\\\\`

Don't forget about the +C at the end.