不定积分的求法

答案是(2/√3)arctan[(2x+1)/√3] + C，第二类换元积分法你学了吗？

x²+x+1 = (x²+x+1/4)+(3/4) = (x)²+2(x)(1/2)+(1/2)²+(3/4) = (x+1/2)²+(√3/2)²
∫ dx/(x²+x+1) = ∫ dx/[(x+1/2)²+(√3/2)²]
Let u = x+1/2，du = dx
=> ∫ du/[u²+(√3/2)²]
Let u = (√3/2)tanz，du = (√3/2)sec²zdz
=> ∫ (√3/2)sec²zdz / [(√3/2)²tan²z+(√3/2)²]
= ∫ (√3/2)sec²zdz / [(√3/2)²(tan²z+1)]
= ∫ sec²zdz / [(√3/2)sec²z] <= 恒等式1+tan²x = sec²x
= [1/(√3/2)]∫ dz
= (2/√3)z + C
= (2/√3)arctan[u/(√3/2)] + C
= (2/√3)arctan[(x+1/2)*(2/√3)] + C
= (2/√3)arctan[(2x+1)/√3] + C

原式=∫dx/[(x+1/2)²+(√3/2)²]
=4/3∫dx/[1+((2x+1)/√3)²]
=2/√3∫d((2x+1)/√3)/[1+((2x+1)/√3)²]
=2arctan((2x+1)/√3)/√3+C