请教大学高数定积分问题
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∫1/(x²-a²)²dx=1/2∫1/[x(x²-a²)²]d(x²-a²)=-1/2∫1/xd1/(x²-a²)
=-1/[2x(x²-a²)]+1/2∫1/(x²-a²)d1/x
=-1/[2x(x²-a²)]-1/2∫1/[x²(x²-a²)]dx
=-1/[2x(x²-a²)]-1/(2a²)∫1/(x²-a²)-1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)∫1/(x-a)-1/(x+a)dx+1/(2a²)∫1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)ln|(x-a)/(x+a)|-1/(2a²x)+C
=-1/[2x(x²-a²)]+1/2∫1/(x²-a²)d1/x
=-1/[2x(x²-a²)]-1/2∫1/[x²(x²-a²)]dx
=-1/[2x(x²-a²)]-1/(2a²)∫1/(x²-a²)-1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)∫1/(x-a)-1/(x+a)dx+1/(2a²)∫1/x²dx
=-1/[2x(x²-a²)]-1/(4a³)ln|(x-a)/(x+a)|-1/(2a²x)+C
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