Question

뜨-3)20),(r-1)3

Answer 1

The given integration is,

∫ ( x−3 ) ( x−1 ) 3 e x dx = ∫ ( x−1−2 ) ( x−1 ) 3 e x dx = ∫ ( ( x−1 ) ( x−1 ) 3 − 2 ( x−1 ) 3 ) e x dx = ∫ ( 1 ( x−1 ) 2 − 2 ( x−1 ) 3 ) e x dx

The given function is in the form of,

∫ e x ( f( x )+ f ′ ( x ) )dx= e x f( x )+C

Assume, f( x )= 1 ( x−1 ) 2 and f ′ ( x )= −2 ( x−1 ) 3 .

Thus, integration of given function is ∫ ( 1 ( x−1 ) 2 − 2 ( x−1 ) 3 ) e x dx= e x ( x−1 ) 2 +C.