题目

求不定积分,用换元法
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

提问时间：2020-08-07

答案

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