Integral (2x^2+4x-3)^5(8x+8) Tolong bantu buat bsok

Integral Substitusi.

∫ (2x² + 4x - 3)⁵ (8x + 8) dx
= 8 ∫ (x + 1)(2x² + 4x - 3)⁵ dx
u = 2x² + 4x - 3
du / dx = 4x + 4
= 8 ∫ (x + 1)u⁵ du / (4x + 4)
= 8(1/4) ∫ u⁵ du
= 2(1/6 u⁶) + C
= 1/3 (2x² + 4x - 3) + C