原式=∫dx/x(x-4)
=1/4∫[1/(x-4)-1/x]dx
=ln|x-4|/4-ln|x|/4+c
∫dx/√(x^2-4x) x=2+2sect
1个回答
相关问题
-
∫[(3x+1)/(4+x^2)^(1/2)]dx ∫1/[(x^2)(4-x^2)]dx
-
∫(x^4+2x^2+4x+1)/( x^2+1)^3}dx 和 ∫(2x^2-12x+4)/(x^3-4x^2)dx,
-
1.∫x/根号下（1+x) dx 2.∫x^2e^x dx 3.∫xarctanx dx 4.∫xsin2x dx
-
不定积分题(9题)1.∫ln(x+1)dx2.∫(√x+1/√x)^2dx3.∫√(1-2x)dx4.∫2^(2x)dx
-
设X=Rcost,y=Rsint,求d2y／dx2 设x=ln∣sect+tant∣,y=ln∣cswt-cott∣求d
-
∫x-根号下x dx ∫lx-2l dx ∫1/根号下（4-x^2) dx ∫e^(-x) dx ∫2/根号下x dx
-
∫(x-2)√(x^2 4x 1) dx
-
为什么∫x√4-x^2dx=1/2∫√4-x^2d（4-x^2﹚?
-
积分∫(x^2)/(x^2+4)dx
-
∫x^2/(9+4x^2)dx