Question

不定积分∫1/x(√2x+1)dx

写下详细过程

Answer 1

t=(√2x+1)，x=1/2*t^2-1/2,dx=tdt
原积分=S1/(1/2*t^2-1/2)t *tdt
=S2/(t^2-1) dt
=S(1/(t-1)-1/(t+1))dt
=ln(t-1)-ln(t+1)+c
=ln(t-1)/(t+1)+c
=ln((√2x+1)-1)/((√2x+1)+1)+c

Answer 2

∫dx/(x√2x+1)
=∫dx/√x^3√(2+1/x)
=(-2)∫d(1/√x)/√[2+(1/√x)^2]
=-2∫d(1/√2x)/√[1+(1/√2x)^2]
1/√2x=tanu secu^2=1/2x+1 cosu^2=2x/(1+2x) sinu=√[1/(1+2x)]
=-2∫secudu
=-2sinu+C
=-2/√(1+2x)+C