Asked by brian

find the integral of (1+3t^5)^20*t^4dt
Answered by Mgraph
Let z=1+3t^5 then
dz=15t^4dt -->t^4dt=(1/15)dz
The integral of (z^20)(1/15)dz=z^21/315+C=
(1+3t^5)^21/315+C
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