凑微分 求大神∫ dx/(a² + x²)= ∫ dx/[a²(1 + x²/a²)]= (1/a²)∫ dx/(1 + x²/a²)= (1/a²)∫ d(x/a · a)/(1 + x²/a²)= (1/a²)(a)∫ d(x/a)/(1 + x²/a²)= (1/a)∫ d(x/a)/[1 + (x/a

凑微分 求大神∫ dx/(a² + x²)= ∫ dx/[a²(1 + x²/a²)]= (1/a²)∫ dx/(1 + x²/a²)= (1/a²)∫ d(x/a · a)/(1 + x²/a²)= (1/a²)(a)∫ d(x/a)/(1 + x²/a²)= (1/a)∫ d(x/a)/[1 + (x/a
凑微分 求大神
∫ dx/(a² + x²)= ∫ dx/[a²(1 + x²/a²)]= (1/a²)∫ dx/(1 + x²/a²)= (1/a²)∫ d(x/a · a)/(1 + x²/a²)= (1/a²)(a)∫ d(x/a)/(1 + x²/a²)= (1/a)∫ d(x/a)/[1 + (x/a)²]= (1/a)arctan(x/a) + C是不是d后边的常数都能直接提到前边?就像题中的a,被提到前边是不是不用考虑微分什么的?（不要给我讲别的方法,我就问的是这种）解答的好有附加分

凑微分 求大神∫ dx/(a² + x²)= ∫ dx/[a²(1 + x²/a²)]= (1/a²)∫ dx/(1 + x²/a²)= (1/a²)∫ d(x/a · a)/(1 + x²/a²)= (1/a²)(a)∫ d(x/a)/(1 + x²/a²)= (1/a)∫ d(x/a)/[1 + (x/a
是的,d后边的常数都能直接提到前边,
因为c是常数的话,d(cx)=cdx,∫cdx=c∫dx