How do you integrate $36/(2x+1)^3$ ?

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Answer 1 · 2018-05-21T19:24:52

$-9/(2x + 1)^2 + C$

$u = 2x + 1 Rightarrow du = 2 cdot dx$

$int 36/u^3 cdot {du}/2 = 18 int u^{-3} cdot du$

$= 18 u^{-2}/{-2} + C = -9/u^2 + C$

How do you integrate 36/(2x+1)^336/(2x+1)^3?