∫0+∞x/(1+x)3dx
∫0+∞[x/(1+x)3]dx=limb→+∞∫0b(1+x-1)/(1+x)3dx =limb→+∞[∫0b[1/(1+x2)]d(1+x)-∫0b[1/(1+x)3]d(1+x)] =limb→+∞[-[1/(1+x)]|0b+1/2•1/(1+x)2|0b =limb→+∞{1-1/(1+b)+1/2•[1/(1+b)2]-1/2}=1/2
∫0+∞x/(1+x)3dx
∫0+∞[x/(1+x)3]dx=limb→+∞∫0b(1+x-1)/(1+x)3dx =limb→+∞[∫0b[1/(1+x2)]d(1+x)-∫0b[1/(1+x)3]d(1+x)] =limb→+∞[-[1/(1+x)]|0b+1/2•1/(1+x)2|0b =limb→+∞{1-1/(1+b)+1/2•[1/(1+b)2]-1/2}=1/2
