不定积分换元法

令√(1+t)=u，得t=u?-1，dt=2udu

∫1/[1+√(1+t)]dt

=∫2u/(1+u)du

=2∫[(1+u)-1]/(1+u)du

=2∫du-2∫1/(1+u)d(1+u)

=2u-2ln(1+u)+C

=2√(1+t)-2ln[1+√(1+t)]+C

令√(x?+a?)=t，得x?=t?-a?，dx?=2tdt

∫√(x?+a?)/xdx

=∫x√(x?+a?)/x?dx

=[∫√(x?+a?)/x?dx?]/2

=[∫2t?t/(t?-a?)dt]/2

=∫[(t?-a?)+a?]/(t?-a?)dt

=∫dt+a?∫1/(t?-a?)dt

=t+aln[(t-a)/(t+a)]/2+C

=√(x?+a?)+aln{[√(x?+a?)-a]/[√(x?+a?)+a]}/2+C

令√(1+2/x)=u，得x=2/(u?-1)，dx=-4u/(u?-1)?

∫√(x?+2x)/x?dx

=∫√[4/(u?-1)?+4/(u?-1)]/[4/(u?-1)?]?[-4u/(u?-1)?]du

=-∫u√[4+4(u?-1)/(u?-1)?]du

=-2∫u?/(u?-1)du

=-2∫[(u?-1)+1]/(u?-1)du

=-2∫du-2∫1/(u?-1)du

=ln[(1+u)/(1-u)]-2u+C

=ln[(1+√(1+2/x))/(1-√(1+2/x))]-2√(1+2/x)+C

此处√(1+2/x)=u经过两次代换

首先令x=1/t，得dx=-1/t?dt，得到∫√(1+2t)/tdt，再令√(1+2t)=u，即√(1+2/x)=u

令√(e^u+1)=t，得u=ln(t?-1)，du=2t/(t?-1)dt

∫1/√(e^u+1)du

=∫1/t?2t/(t?-1)dt

=∫1/(t?-1)dt

=ln[(t-1)/(t+1)]+C

=ln[(√(e^u+1)-1)/(√(e^u+1)+1)]+C

令x=1/t，得dx=-1/t?dt

∫1/x√(a?-b?x?)dx

=-∫t/√(a?-b?/t?)?1/t?dt

=-∫t?/√(a?t?-b?)?1/t?dt

=-∫1/a√[t?-(b/a)?]dt

=-ln[t+√(t?-b?/a?)]/a+C

=-ln[1/x+√(1/x?-b?/a?)]/a+C

=ln{ax/[a+√(a?-b?x?)]}/a+C

令√(1+lnx)=t，得x=e^(t?-1)，dx=2te^(t?-1)

∫√(1+lnx)/xlnxdx

=∫t/(t?-1)e^(t?-1)?2te^(t?-1)dt

=2∫t?/(t?-1)dt

=2∫[(t?-1)+1]/(t?-1)dt

=2∫dt+2∫1/(t?-1)dt

=2t+ln[(t-1)/(t+1)]+C

=2√(1+lnx)+ln[(√(1+lnx)-1)/(√(1+lnx)+1)]+C