不定积分∫2^x/（1+2^x+4^x)dx

Question

不定积分 ∫ 2^x /（1+2^x+4^x) dx

Answer 1

请耐心看完~
令t = 2^x,dt = 2^x ln2 dx
∫ 2^x/(1 + 2^x + 4^x) dx
= ∫ t/(1 + t + t²) * 1/(t ln2) dt
= 1/ln2 ∫ dt/[(t + 1/2)² + 3/4]
令t + 1/2 = √(3/4) tanz,dt = √3/2 sec²z dz
= 1/ln2 ∫ (√3/2 sec²z)/(3/4 tan²z + 3/4) dz
= (1/ln2)(√3/2)(4/3)∫ sec²z/sec²z dz
= 2/(√3 ln2) z + C
= 2/(√3 ln2) arctan[(t + 1/2)/√(3/4)] + C
= 2/(√3 ln2) arctan[(2t + 1)/√3] + C
= 2/(√3 ln2) arctan{[2^(x + 1) + 1]/√3} + C