∫1/[x(1+√x)^2dx∫1/[x(1+√x)^2]dx
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![∫1/[x(1+√x)^2dx∫1/[x(1+√x)^2]dx](/uploads/image/z/15160743-63-3.jpg?t=%E2%88%AB1%2F%5Bx%281%2B%E2%88%9Ax%29%5E2dx%E2%88%AB1%2F%5Bx%281%2B%E2%88%9Ax%29%5E2%5Ddx)
∫1/[x(1+√x)^2dx∫1/[x(1+√x)^2]dx
∫1/[x(1+√x)^2dx
∫1/[x(1+√x)^2]dx
∫1/[x(1+√x)^2dx∫1/[x(1+√x)^2]dx
原式=∫2√xd(√x)/[(√x)²(1+√x)²]
=2∫d(√x)/[√x(1+√x)²]
=2∫[1/√x-1/(1+√x)-1/(1+√x)²]d(√x)
=2[ln(√x)-ln(1+√x)+1/(1+√x)]+C (C是积分常数)
=2[ln(√x/(1+√x))+1/(1+√x)]+C.