Question

求问两题不定积分的题目！！

Answer 1

解：1.原式=∫(2/(1+x)-1)²e^xdx
=∫[4/(1+x)²-4/(1+x)+1]e^xdx
=4∫e^xdx/(1+x)²-4∫e^xdx/(1+x)+∫e^xdx
=4[-e^x/(1+x)+∫e^xdx/(1+x)]-4∫e^xdx/(1+x)+e^x (第一个积分应用分部积分法)
=-4e^x/(1+x)+4∫e^xdx/(1+x)]-4∫e^xdx/(1+x)+e^x+C (C是积分常数)
=-4e^x/(1+x)+e^x+C；
2.原式=-lnx/√(x²-1)+∫dx/[x√(x²-1)] (应用分部积分法)
=-lnx/√(x²-1)+(1/2)∫d(x²)/[x²√(x²-1)]
=-lnx/√(x²-1)+(1/2)∫2tdt/[(t²+1)t] (令√(x²-1)=t，则x²=t²+1，d(x²)=2tdt)
=-lnx/√(x²-1)+∫dt/(t²+1)
=-lnx/√(x²-1)+arctant+C (C是积分常数)
=-lnx/√(x²-1)+arctan[√(x²-1)]+C。

Answer 2

1
∫[(1-x)/(1+x)]^2 e^xdx
=∫[(1-x)/(1+x)]^2de^x [(1-x)/(1+x)]=[-1+2/(1+x)]
=[(1-x)/(1+x)]^2 *e^x -∫e^xd[(1-x)/(1+x)]^2
=[(1-x)/(1+x)]^2 *e^x -∫e^x *2*[-1+2/(1+x)]*(-2)dx/(1+x)^2
=[(1-x)/(1+x)]^2 e^x -4∫e^xdx/(1+x)^2+4∫e^x*2dx/(1+x)^3
=[(1-x)/(1+x)]^2 e^x-4[ ∫de^x/(1+x)^2+∫e^xd(1/(1+x)^2)]
=e^x [(1-x)/(1+x)]^2 -4[e^x/(1+x)^2- ∫e^xd(1/(1+x^2))+∫e^xd(1/(1+x)^2)]+c
=e^x[(1-x)/(1+x)]^2-4e^x/(1+x)^2+C

2
∫xlnxdx/(x^2-1)^(3/2)
x=secu
=∫seculnsecudsecu/tanu^3
=∫secu^2lnsecudu/tanu^2
=∫lnsecudtanu/tanu^2
=-∫lnsecud(1/tanu)
=-lnsecu /tanu +∫(1/tanu)dlnsecu
=(-lnsecu)/tanu+ ∫secudu/secu
=(-lnsecu)/tanu+u+C
=(-lnx)/√(x^2-1)+arccos(1/x)+C