求不定积分，用换元法

求不定积分，用换元法

令√(1+t)=u，得t=u²-1，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√(x²+a²)=t，得x²=t²-a²，dx²=2tdt∫√(x²+a²)/xdx=∫x√(x²+a²)/x²dx=[∫√(x²+a²)/x²dx²]/2=[∫2t•t/(t²-a²)dt]/2=∫[(t²-a²)+a²]/(t²-a²)dt=∫dt+a²∫1/(t²-a²)dt=t+aln[(t-a)/(t+a)]/2+C=√(x²+a²)+aln{[√(x²+a²)-a]/[√(x²+a²)+a]}/2+C令√(1+2/x)=u，得x=2/(u²-1)，dx=-4u/(u²-1)²∫√(x²+2x)/x²dx=∫√[4/(u²-1)²+4/(u²-1)]/[4/(u²-1)²]•[-4u/(u²-1)²]du=-∫u√[4+4(u²-1)/(u²-1)²]du=-2∫u²/(u²-1)du=-2∫[(u²-1)+1]/(u²-1)du=-2∫du-2∫1/(u²-1)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/t²dt，得到∫√(1+2t)/tdt，再令√(1+2t)=u，即√(1+2/x)=u令√(e^u+1)=t，得u=ln(t²-1)，du=2t/(t²-1)dt∫1/√(e^u+1)du=∫1/t•2t/(t²-1)dt=∫1/(t²-1)dt=ln[(t-1)/(t+1)]+C=ln[(√(e^u+1)-1)/(√(e^u+1)+1)]+C令x=1/t，得dx=-1/t²dt∫1/x√(a²-b²x²)dx=-∫t/√(a²-b²/t²)•1/t²dt=-∫t²/√(a²t²-b²)•1/t²dt=-∫1/a√[t²-(b/a)²]dt=-ln[t+√(t²-b²/a²)]/a+C=-ln[1/x+√(1/x²-b²/a²)]/a+C=ln{ax/[a+√(a²-b²x²)]}/a+C令√(1+lnx)=t，得x=e^(t²-1)，dx=2te^(t²-1)∫√(1+lnx)/xlnxdx=∫t/(t²-1)e^(t²-1)•2te^(t²-1)dt=2∫t²/(t²-1)dt=2∫[(t²-1)+1]/(t²-1)dt=2∫dt+2∫1/(t²-1)dt=2t+ln[(t-1)/(t+1)]+C=2√(1+lnx)+ln[(√(1+lnx)-1)/(√(1+lnx)+1)]+C