bantu jawab. menggunakan integral substitusi....

tidak semua bentuk integral subtitusi

1) ∫ 8x²+4x-3 dx = ⁸/₃ x³ + 2x² - 3x ∑²₀
                           = [ ⁸/₃ (2)³ + 2(2)² - 3(2)] - [ 0 ]
                           = ⁶⁴/₃ + 8 - 6
                           = ⁷⁰/₃

2) ∫ 3 cos x + 2 sin x dx  = 3 sin x + (-2 cos x) Σ²ⁿ₀
                                         = [3 sin 2π - 2 cos 2π] - [3 sin 0 - 2 cos 0]
                                         = [ 0 - 2] - [0 - 2]
                                         = -2 - (-2)
                                         = 0

 
  3)     mis : u = x⁴+ 1 --> du = 3x³ dx
                                     dx = du/ 3x³
 
       ∫ 8x³ (x⁴+1)⁷ dx  = ∫ 8x³ u⁷ .du/ 3x³
                                  = ⁸/₃ ∫ u⁷ du
                                  = ⁸/₃. ¹/₈. u⁸
                                  = ¹/₃ u⁸
                                  = ¹/₃ (x⁴+1) ∑¹₀
                                  = ¹/₃ [ (1⁴+1) -(0+1) ]
                                  = ¹/₃ [(2)-(1)]
                                  = ¹/₃