What is #int (x+5)/sqrt(9-(x-3)^2) dx#?

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Answer 1 · 2018-05-22T13:17:32

See below

Lets make the change #x-3=3sint# with this change

#dx=3costdt# and #x+5=3sint+8#

#I=int(3sint+8)/sqrt(3^2-3^2sin^2t)·3costdt=#

#=int((3sint+8)(3costdt))/(3(sqrt(1-sin^2t))=#

#=int3(sint+8)dt=-3cost+24t+C=#

With change made #sint=(x-3)/3# and #sqrt(1-(x-3)^2/9)=cost=#

#sqrt(9-x^2+6x-9)/3=cost# and #t=arcsin((x-3)/3)#

Finally we have #I=-sqrt(6x-x^2)+24arcsin((x-3)/3)+C#