Dear student We have ∫(x-4)ex(x-2)3dxhere let (x-2)=u⇒x=u+2⇒dx=du⇒∫(x-4)ex(x-2)3dx=∫(u-2)eu+2u3du⇒∫(x-4)ex(x-2)3dx=∫ueue2u3du-∫2eue2u3du⇒∫(x-4)ex(x-2)3dx=e2∫euu2du-2e2∫euu3dunow we have the integral∫exx-tdx=(-x)tx-tΓ(1-t,-x)+Cwhere Γ(a,x) is the incomplete Gamma function Thus we have ∫(x-4)ex(x-2)3dx=e2-u2(u)-2Γ(1-2,-u)-2e2-u3(u)-3Γ(1-3,-u)and Γ(n)=(n-1)!Thus on simplification and putting back x-2=u we get ∫(x-4)ex(x-2)3dx=ex(x-2)2+C Regards

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W e h a v e \int \frac{(x - 4) e^{x}}{(x - 2)^{3}} d x h e r e l e t (x - 2) = u \Rightarrow x = u + 2 \Rightarrow d x = d u \Rightarrow \int \frac{(x - 4) e^{x}}{(x - 2)^{3}} d x = \int \frac{(u - 2) e^{u + 2}}{u^{3}} d u \Rightarrow \int \frac{(x - 4) e^{x}}{(x - 2)^{3}} d x = \int \frac{u e^{u} e^{2}}{u^{3}} d u - \int \frac{2 e^{u} e^{2}}{u^{3}} d u \Rightarrow \int \frac{(x - 4) e^{x}}{(x - 2)^{3}} d x = e^{2} \int \frac{e^{u}}{u^{2}} d u - 2 e^{2} \int \frac{e^{u}}{u^{3}} d u n o w w e h a v e t h e i n t e g r a l \int e^{x} x^{- t} d x = (- x)^{t} x^{- t} Γ (1 - t, - x) + C w h e r e Γ (a, x) i s t h e i n c o m p l e t e G a m m a f u n c t i o n T h u s w e h a v e \int \frac{(x - 4) e^{x}}{(x - 2)^{3}} d x = e^{2} {(- u)}^{2} (u)^{- 2} Γ (1 - 2, - u) - 2 e^{2} {(- u)}^{3} (u)^{- 3} Γ (1 - 3, - u) a n d Γ (n) = (n - 1)! T h u s o n s i m p l i f i c a t i o n a n d p u t t i n g b a c k x - 2 = u w e g e t \int \frac{(x - 4) e^{x}}{(x - 2)^{3}} d x = \frac{e^{x}}{(x - 2)^{2}} + C

Regards

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