I'm unsure of how to go about this problem. If someone can please help and explain it to me step by step I would be so appreciative. Thank you in advance!
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∫(4x - 9)^(-3) dx
u = 4x - 9
du/4 = dx
1/4*∫u^(-3) du
-1/8*(4x - 9)^(-2) + C
u = 4x - 9
du/4 = dx
1/4*∫u^(-3) du
-1/8*(4x - 9)^(-2) + C
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Integrate using the power rule: -(1/2) * (4x - 9)^(-2) * (1/4) + c
The power rule states raise the exponent by 1 then divide by the new number. The second part is the chain rule which is the division of the derivative.
Simplify: -(1/8) * (4x - 9)^(-2) + c
Simplify: - (1 / (8 * (4x - 9)^2)) +c
The power rule states raise the exponent by 1 then divide by the new number. The second part is the chain rule which is the division of the derivative.
Simplify: -(1/8) * (4x - 9)^(-2) + c
Simplify: - (1 / (8 * (4x - 9)^2)) +c
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The easiest way is to differentiate (4x-9)^(-2)
You get -8(4x-9)-3
So the integral is -(1/8)(4x-9)^(-2)
People always seem to do these in a far too complicated way!
You get -8(4x-9)-3
So the integral is -(1/8)(4x-9)^(-2)
People always seem to do these in a far too complicated way!
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∫ dx/(4x - 9)^3
Let u = 4x - 9
then du = 4 dx
and 1/4 du = dx
1/4 ∫ du/u^3 = 1/4 ∫ u^(-3) du = 1/4 u^(-2)/(-2) = -1/8 u^(-2) = -1/[8(4x - 9)^2] + C
Let u = 4x - 9
then du = 4 dx
and 1/4 du = dx
1/4 ∫ du/u^3 = 1/4 ∫ u^(-3) du = 1/4 u^(-2)/(-2) = -1/8 u^(-2) = -1/[8(4x - 9)^2] + C