Question

求不定积分 谢谢

Answer 1

let
u =e^(x/6)
du = (1/6) e^(x/6) dx
dx =6du/u
∫dx/[1+e^(x/2)+e^(x/3)+e^(x/6)]
=6∫du/[u(1+u+u^2+u^3)]
=6∫ (u-1)/[u(u^4 -1)] du
=3∫ [(u-1)/u] [1/(u^2-1) - 1/(u^2 +1) ] du
=(3/2)∫ [(u-1)/u] . [ 1/(u-1) -1/(u+1) - 2/(u^2 +1) ] du
=(3/2)∫ ( 1 - 1/u) . [ 1/(u-1) -1/(u+1) - 2/(u^2 +1) ] du
=(3/2)[ ln|(u-1)/(u+1)| -2arctanu]
-(3/2) ∫ (1/u) . [ 1/(u-1) -1/(u+1) - 2/(u^2 +1) ] du
=(3/2)[ ln|(u-1)/(u+1)| -2arctanu] -(3/2)ln|(u-1)(u+1)/u^2| - ln|u^2+1| + C
where u = e^(x/6)
consider
∫ (1/u) . [ 1/(u-1) -1/(u+1) - 2/(u^2 +1) ] du
=∫ { [1/(u-1) -1/u] - [1/u -1/(u+1) ] } du -2∫ du/[u(u^2 +1)] du
= ln|(u-1)(u+1)/u^2| - ln|u^2+1| + C'

Answer 2

∫lnxdx/x
=∫lnxdlnx
=(1/2)(lnx)^2+c

∫sinx(cosx)^3dx
=-∫(cosx)^3dcosx
=-(1/4)cos^4x+c

∫arctgxdx
=xarctgx-∫x/(1+x^2)dx
=xarctgx-(1/2)∫d(x^2+1)/(x^2+1)
=xarctgx-(1/2)ln(x^2+1)+c