求不定积分 1.ʃx^4/(x²+1)dx 2.ʃ（1+cos²x)/(1+cos2x）dx

∫ x⁴/(x² + 1) dx
= ∫ x²(x² + 1 - 1)/(x² + 1) dx
= ∫ x² dx - ∫ (x² + 1 - 1)/(x² + 1) dx
= x³/3 - ∫ dx + ∫ dx/(x² + 1)
= x³/3 - x + arctan(x) + C
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∫ (1 + cos²x)/(1 + cos2x) dx
= ∫ [1 + (1 + cos2x)/2]/(1 + cos2x) dx
= (1/2)∫ (2 + 1 + cos2x)/(1 + cos2x) dx
= (1/2)∫ dx + ∫ dx/(1 + cos2x)
= x/2 + ∫ (1 - cos2x)/[(1 + cos2x)(1 - cos2x)] dx
= x/2 + ∫ (1 - cos2x)/sin²2x dx
= x/2 + (1/2)∫ (csc²2x - csc2xcot2x) d(2x)
= x/2 + (1/2)(1 - cos2x)/sin2x + C
= x/2 + (1/2)[1 - (1 - 2sin²x)]/(2sinxcosx) + C
= x/2 + (1/2)(2sinx)/(2cosx) + C
= (1/2)(x + tanx) + C

1、 =∫(x^4+x²-x²-1+1)/(x²+1)dx=∫[x²-1+1/(x²+1)]dx=(1/3)x³-x+arctanx+C
2、 =∫(1+cos²x)/(2cos²x)dx=(1/2)∫[1+1/cos²x]dx=(1/2)x+(1/2)tanx+C