求不定积分，用换元法

Question

1）∫1/(1+根号(1+t))dt

2)∫根号(x^2+a^2)/xdx

3)∫根号(x^2+2x)/x^2dx

4)∫1/根号(e^u+1)du

5)∫1/x*根号(a^2-b^2*x^2)dx

6)∫根号(1+lnx)/x*lnxdx

请写出过程 谢谢

Answer 1

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本回答被提问者采纳