第1个回答 2014-12-18
(5)解:∫ x²sin2x dx = x²[-0.5cos2x] - ∫ [-0.5cos2x] dx²
= -0.5x²cos2x + ∫ xcos2x dx
= -0.5x²cos2x + { x[0.5sin2x] - ∫ [0.5sin2x] dx }
= -0.5x²cos2x + { 0.5xsin2x - 0.5 ∫ sin2x dx }
= -0.5x²cos2x + 0.5xsin2x +0.25cos2x + C (C为任意常数)
(7)解:∫ ln(x+√(1+x²)) dx
= xln(x+√(1+x²)) - ∫ x dln(x+√(1+x²))
= xln(x+√(1+x²)) - ∫ x [1/√(1+x²)] dx
= xln(x+√(1+x²)) - √(1+x²) + C (C为任意常数)