Integral of 3^sqrt(4x - 3)

[@alexchan4226]

+x/2

[@gelbkehlchen]

Solution:
∫3^[√(4x-3)]*dx = ∫e^❮ln{3^[√(4x-3)]}❯*dx = ∫e^{ln(3)*[√(4x-3)]}*dx =
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Substitution:
u = √(4x-3) du = 1/2*1/√(4x-3)*4*dx dx = √(4x-3)/2*du = u/2*du
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= 1/2*∫u*e^[ln(3)*u]*du =
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Solution by partial integration:
Partial integration can be derived from the product rule of differential calculus. The product rule of differential calculus states:
(u*v)’ = u’*v+u*v’ |-u’*v ⟹
(u*v)’-u’*v = u*v’ ⟹
u*v’ = (u*v)’-u’*v |∫() ⟹
∫u*v’*dx = u*v-∫u’*v*dx
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= 1/2*{u*e^[ln(3)*u]/ln(3)-1/ln(3)*∫e^[ln(3)*u]*du}
= 1/2*{u*e^[ln(3)*u]/ln(3)-1/ln²(3)*e^[ln(3)*u]+C}
= 1/[2*ln(3)]*[u*3^u-3^u/ln(3)]+C/2
= 3^u/ln(3²)*[u-1/ln(3)]+D
= 3^[√(4x-3)]/ln(9)*[√(4x-3)-1/ln(3)]+D