分别求x(1+x^2)和（sinx*cosx）/(sinx+cosx)两个式子的不定积分.

第一题没搞错题吧?
∫ x(1+x²)dx
=∫ (x+x³)dx
=(1/2)x²+(1/4)x^4+C
∫ (sinx*cosx）/(sinx+cosx)dx
=(1/2)∫ (2sinx*cosx）/(sinx+cosx)dx
=(1/2)∫ (2sinx*cosx+1-1）/(sinx+cosx)dx
=(1/2)∫ (2sinx*cosx+1）/(sinx+cosx)dx-(1/2)∫ 1/(sinx+cosx)dx
=(1/2)∫ (2sinx*cosx+sin²x+cos²x）/(sinx+cosx)dx-(1/2)∫ 1/(sinx+cosx)dx
=(1/2)∫ (sinx+cosx)²/(sinx+cosx)dx-(1/(2√2))∫ 1/((1/√2)sinx+(1/√2)cosx)dx
=(1/2)∫ (sinx+cosx)dx-(1/(2√2))∫ 1/sin(x+π/4)dx
=(1/2)(sinx-cosx)-(1/(2√2))ln|csc(x+π/4)-cot(x+π/4)|+C