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a> bisa dkalikan dulu
b> dengan integral parsial = v du + u dv
c>caea substitusi
cara c> substitusi
[x+2](x² + 2)⁵ = x(x² + 2)⁵+ 2(x² + 2)⁵
untuk yan belakang hasil integral = 2/6 (x² + 2)^6 . 2x
untuk yang kiri
misal kan u=(x² + 2) ......du = 2x dx..
x(x² + 2)⁵ dx = u . 1/2. du = 1/4 u² = 1/4[(x² + 2)².
hasil integral = 2/6(x² + 2)^6 + 1/4 (x² + 2)² + C