Question

dx／（x^2*（ax+b）^0.5）的不定积分，要详解'我设（ax+b）^0.5=u没做出来'请...

dx／（x^2*（ax+b）^0.5）的不定积分，要详解'我设（ax+b）^0.5=u没做出来'请大侠指导
那个是根号下啊'晕死答案不对

Answer 1

∫dx/x^2√(ax+b)
=(-1/√b)∫d(1/x)/∫√(ax/b+1)
=(-a/b√b)∫d(b/ax)/√(ax/b+1)
b/ax=tanu^2
=(-a/b√b)∫2tanusecu^2 du/sinu
=(-2a/b√b)∫secu^3du
=(-a/b√b)[secutanu+ln|secu+tanu|
=(-a/b√b)[√(b/ax)+1) *(b/ax)+ln|√[(b/ax)+1] +(b/ax)|+C

∫secx^3dx=∫secxdtanx=secxtanx-∫tanx^2secxdx=secxtanx-∫(secx)^3dx+∫secxdx
2∫secx^3dx=secxtanx+∫secxdx=secxtanx+ln|secx+tanx|

追问

答案不一样啊

追答

答案? 比较一下