Question

换元法求不定积分1/根号(x^2+2x+5)dx

有过程谢谢

Answer 1

原式=∫1/√[(x+1)²+4]d(x+1)
设x+1=2tant，t=actan[(x+1)/2]，
则√[(x+1)²+4]=√[4(tan²t+1)]=√(4sec²t)=2sect，d(x+1)=2sec²tdt
∴原式=∫1/√[(x+1)²+4]d(x+1)
=∫1/(2sect)*2sec²tdt
=∫sectdt
=ln|sect+tant|+C
=ln|sec(actan[(x+1)/2])+[(x+1)/2]|+C
=ln|√(1+[(x+1)/2]²)+[(x+1)/2])|+C

Answer 2

原式=∫1/[(x+1)²+4]d(x+1)
=1/2arctan(x+1)/2+C