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∫1/(x²+1)²dx?
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∫1/(x²+1)²dx
=(1/2)∫((1-x²)+(1+x²))/(x²+1)²dx
=(1/2)∫(1-x²)/(x²+1)²dx+(1/2)∫1/(x²+1)dx
=(1/2)∫(1·(1+x²)-x·2x)/(x²+1)²dx+(1/2)∫1/(x²+1)dx
=(1/2)·x/(x²+1)+(1/2)arctanx+C
=x/(2x²+2)+(1/2)arctanx+C
=(1/2)∫((1-x²)+(1+x²))/(x²+1)²dx
=(1/2)∫(1-x²)/(x²+1)²dx+(1/2)∫1/(x²+1)dx
=(1/2)∫(1·(1+x²)-x·2x)/(x²+1)²dx+(1/2)∫1/(x²+1)dx
=(1/2)·x/(x²+1)+(1/2)arctanx+C
=x/(2x²+2)+(1/2)arctanx+C
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