(√x^2+1)/x的不定积分和(√x^2+1)/x^2的不定积分

Question

(√x^2+1)/x的不定积分和(√x^2+1)/x^2的不定积分

Answer 1

设x=tant
则dx=（sect）^2dt,
(√x^2+1)/x的不定积分
=不定积分[(sect)^3/tant]dt
=不定积分{sint/[(cost)^2(sint)^2]}dt
= -不定积分{1/[(1-(cost)^2)(cost)^2]}dcost
=不定积分{1/[cost)^2-1]-(1/cost)^2}dcost
=1/2不定积分[(1/(1-cost)+(1/(cost+1)]dcost-不定积分(1/cost)^2dcost
=ln[(√x^2+1)/x]+(√x^2+1)/[3(1+x^2)^2]+c
(√x^2+1)/x^2的不定积分
=不定积分[(sect)^3/(tant)^2]dt
=不定积分{cost/[(cost)^2(sint)^2]}dt
=不定积分{1/[(1-(sint)^2)(sint)^2]}dsint
=不定积分{1/[(1-(sint)^2)+1/(sint)^2]}dsint
=不定积分{(1/2)[1/(1-sint)+1/(1+sint)]+1/(sint)^2]}dsint
=-(1/2)ln(1-sint)+(1/2)ln(1+sint)-1/[3(sint)^3+c
=ln(x+√x^2+1)-(1+x^2)(√x^2+1)/(3x^3)+c