简单的高数，不定积分题目，换元法，求数学帝来帮帮忙！谢了

Question

如题所述

Answer 1

1、令x=1/t dx=-dt/t^2
原式=-∫tdt/√(t^4+1)
=-1/2*∫d(t^2)/√[(t^2)^2+1]
=-1/2*ln|t^2+√(t^4+1)|+C
=-1/2*ln|1/x^2+√(1/x^4+1)|+C
2、令x=sint dx=costdt
原式=∫costdt/(sint+cost)
令A=∫costdt/(sint+cost) B=∫sintdt/(sint+cost)
A+B=∫(sint+cost)dt/(sint+cost)=t+C1
A-B=∫(cost-sint)dt/(sint+cost)=∫d(sint+cost)/(sint+cost)=ln|sint+cost|+C2
A=(t+ln|sint+cost|)/2+C
所以原式=(arcsinx+ln|x+√(1-x^2)|)/2+C
3、原式=∫[1+√x-√(1+x)]dx/(1+2√x+x-1-x)
=1/2*∫x^(-1/2)+1-√(1+x)/√x dx
=√x+x/2-1/2*∫√(1+x)dx/√x
令t=√x t^2=x 1+x=t^2+1 dx=2tdt
原式=√x+x/2-∫√(t^2+1)dt
=√x+x/2-t/2*√(t^2+1)-1/2*ln|t+√(t^2+1)|+C
=√x+x/2-1/2*√(x^2+x)-1/2*ln|√x+√(x+1)|+C
4、令t=lnx x=e^t dx=e^tdt
原式=∫√(1+t)dt/t
再令u=√(1+t) t=u^2-1 dt=2udu
原式=2*∫u^2du/(u^2-1)
=2*∫1+1/(u^2-1) du
=2*∫du+∫du/(u-1)-∫du/(u+1)
=2u+ln|u-1|-ln|u+1|+C
=2√(1+t)+ln|√(1+t)-1|-ln|√(1+t)+1|+C
=2√(1+lnx)+ln|√(1+lnx)-1|-ln|√(1+lnx)+1|+C

Answer 2

一、令x=1/t dx=-dt/t^2
原式=-∫tdt/√(t^4+1)
=-1/2*∫d(t^2)/√[(t^2)^2+1]
=-1/2*ln|t^2+√(t^4+1)|+C
=-1/2*ln|1/x^2+√(1/x^4+1)|+C
二、令x=sint dx=costdt
原式=∫costdt/(sint+cost)
令A=∫costdt/(sint+cost) B=∫sintdt/(sint+cost)
A+B=∫(sint+cost)dt/(sint+cost)=t+C1
A-B=∫(cost-sint)dt/(sint+cost)=∫d(sint+cost)/(sint+cost)=ln|sint+cost|+C2
A=(t+ln|sint+cost|)/2+C
所以原式=(arcsinx+ln|x+√(1-x^2)|)/2+C
三、原式=∫[1+√x-√(1+x)]dx/(1+2√x+x-1-x)
=1/2*∫x^(-1/2)+1-√(1+x)/√x dx
=√x+x/2-1/2*∫√(1+x)dx/√x
令t=√x t^2=x 1+x=t^2+1 dx=2tdt
原式=√x+x/2-∫√(t^2+1)dt
=√x+x/2-t/2*√(t^2+1)-1/2*ln|t+√(t^2+1)|+C
=√x+x/2-1/2*√(x^2+x)-1/2*ln|√x+√(x+1)|+C
四、令t=lnx x=e^t dx=e^tdt
原式=∫√(1+t)dt/t
再令u=√(1+t) t=u^2-1 dt=2udu
原式=2*∫u^2du/(u^2-1)
=2*∫1+1/(u^2-1) du
=2*∫du+∫du/(u-1)-∫du/(u+1)
=2u+ln|u-1|-ln|u+1|+C
=2√(1+t)+ln|√(1+t)-1|-ln|√(1+t)+1|+C
=2√(1+lnx)+ln|√(1+lnx)-1|-ln|√(1+lnx)+1|+C