Question

Mohon bantuanny.integral (8x-20) (x^2-5x+3) ^ 4 dx

Answer 1

Pake integral parsial
v=(x^2-5x+3)
dv/dx= 2x-5
dx=dv/2x-5

=Int (8x-20) . v^4 . dv/(2x-5) -->coret 8x-20 sama 2x-5, sisanya 4
=Int 4. v^4 . dv
= 4/5 . v^5 + c
= 4/5 . (x^2-5x+3)^5 + c