Integrals Using Substitution
I = ∫-2x+1(-4x2+4x+3) dxPut u = -4x2+4x+3, then du = (-8x+4) dxor, (14) du = (-2x+1) dxI = ∫1(u)(14) duI = 14∫1(u) duI = 14ln (u)I = 14ln(-4x2+4x+3)+C
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Copyright © Dayal D. Purohit, Ph.D.(Mathematics)