Question

求∫(1+x)/[(1+x²)²]

Answer 1

∫(1+x)/(1+x^2)^2 dx
令x=tanz,dz=sec^2z dz
分母(1+x^2)^2=(1+tan^2z)^2=sec^4z
原式=∫(1+tanz)/sec^4z*sec^2z dz
=∫(1+tanz)cos^2z dz
=∫(1+tanz)(1-sin^2z) dz
=∫(1+tanz-sin^2z-tanzsin^2z) dz
=∫dz+∫tanz dz-∫sin^2z dz-∫tanz(1-cos^2z) dz
=∫闹宽dz-(1/2)∫(1-cos2z) dz+∫sinzcosz dz
=(1/2)∫ dz+(1/4)∫cos2z d(2z)-∫cosz d(cosz)
=(1/弊弯粗2)z+(1/4)sin2z-(1/2)cosz^2z+C
=(1/2)z+(1/2)sinzcosz-(1/2)cos^2z+C
=(1/2)arctanx+(1/2)*x/√租镇(1+x^2)*1/√(1+x^2)-(1/2)*[1/√(1+x^2)]^2+C
=(1/2)arctanx+(1/2)[x/(1+x^2)]-1/[2(1+x^2)]+C
=(1/2)[arctanx+(x-1)/(1+x^2)]+C