integral (x-1)^5(x+4)dx tolong bantu dong

misal x-1 = U
jadi x+4 = U + 5
dan du/dx = 1 jadi dU = dx

∫(x-1)^5(x+4)dx = ∫U^5(U+5)dU

= ∫ U^6 + 5U^5 du
= 1/7 U^7 + 5/6 U^6
= 1/7(x-1)^7 + 5/6 (x-1)^6

Int (x + 4) (x - 1)^5 dx => kalo kita gunakan integral parsial dengan cara tabel
turunan .... Integral
(+) (x + 4) ....... (x - 1)^5
(-) 1 ................. 1/6 (x - 1)^6
(+) 0 ................ 1/6. 1/7 (x - 1)^7 = 1/42 (x - 1)^7
(x + 4)(1/6 (x - 1)^6) - 1/42 (x - 1)^7 + C
= 1/6 (x + 4) (x - 1)^6 - 1/42 (x - 1)^7 + C

= 7/42 (x + 4) (x - 1)^6 - 1/42 (x - 1)(x - 1)^7 + C
= 1/42 (x - 1)^6 (7(x + 4) - (x - 1)) + C
= 1/42 (x - 1)^6 (6x + 29) + C