换元,令x=tant
x=0,t=0
x=√3,t=π/3
dx=sec^2tdt
∫[0,√3] dx/[(x^2)*√(1+x^2)]
=∫[0,π/3] 1/[(tan^2t)*sect]*sec^2tdt
=∫[0,π/3] 1/[(tan^2t)*cos^2t]*costdt
=∫[0,π/3] 1/(sin^2t)dsint
=1/3sin^3t[0,π/3]
=√3/8
这题积分可能有点问题,分母有x^2.现就不定积分提示:
分子1=(x^2+1)-x^2,第二个约掉x^2后可积分.第一个约掉√(1+x^2)后,√(1+x^2)/x^2*dx=-√(1+x^2)d(1/x).可用分部积分