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 $I=-sqrt(6x-x^2)+24arcsin((x-3)/3)+C$

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