How do you solve the integral 8/(y-9)^3 dy??

Question

I know the answer is -4/(y-9)^2 +c....but i cant figure out how to get there....HELP!!

Answer 1

I = 8 ∫ [ 1 / (y - 9) ³ ] dy
Let u = y - 9
du = dy
I = 8 ∫ u^( - 3 ) du
I = 8 u^( - 2 ) / ( - 2 ) + C
I = - 4 / u ² + C
I = (- 4) / (y - 9)² + C

Answer 2

just apply the standard integrating formula
∫ay^b dy = [a/(b + 1)]y^(b + 1)

8/(y - 9)³ = 8(y - 9)^(-3)

so
∫8(y - 9)^(-3) dy = [8(y-9)/-2]^(-2) + c
= -4(y - 9)^(-2) + c

this is the same as:
-4/(y - 9)² + c

ps as long as the bit in brackets differentiates to 1, you can treat as if it is an easy integral. If not you have to divide by the differential of the brackets.

Answer 3

This is a standard integral of the type
int[ (ax+b)^m dx ] = (1/a) * [ (ax+b)^(m+1) / (m+1) + C

so int [ 8 *(y-9)^(-3) dxy] = 8 * (1/1) * [ (y-9)^(-2) / (-2) + C
= -4/(y-9)^2 +c....

Answer 4

y - 9 = u
dy = du

8(Integral 1/u^3 du)
8((-1/2)(1/u^2)) + C
-4 (1 / (y-9)^2) + C
-4 / (y-9)^2 + C

Answer 5

divide 8 by 9 and do the 3 thingy. idk. im not that smart, im only 13!!! srry to waste your time.