Asked by Anonymous
How do you integrate (1+ square foot of x)^18 dx?
Answers
Answered by
oobleck
let u = 1+√x
u-1 = √x
du = 1/(2√x) dx = 1/(2(u-1)) dx
so dx = 2(u-1) du
now just integrate 2u^18(u-1) du
Post your work if you get stuck. You should wind up with
1/190 (1+√x)^19 (19√x - 1)
u-1 = √x
du = 1/(2√x) dx = 1/(2(u-1)) dx
so dx = 2(u-1) du
now just integrate 2u^18(u-1) du
Post your work if you get stuck. You should wind up with
1/190 (1+√x)^19 (19√x - 1)
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