∫[3x^2/(1+x^2)dx为何是3x-3arctanx+C?∫[3x^2/(1+x^2)dx=∫[3-3/(1+x^2)]dx=3∫[1-1/(1+x^2)]dx=3[(x+c)-(arctanx+c)]=3x-3arctanx为何正确答案后面还有加C呢
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![∫[3x^2/(1+x^2)dx为何是3x-3arctanx+C?∫[3x^2/(1+x^2)dx=∫[3-3/(1+x^2)]dx=3∫[1-1/(1+x^2)]dx=3[(x+c)-(arctanx+c)]=3x-3arctanx为何正确答案后面还有加C呢](/uploads/image/z/9308069-53-9.jpg?t=%E2%88%AB%5B3x%5E2%2F%281%2Bx%5E2%29dx%E4%B8%BA%E4%BD%95%E6%98%AF3x-3arctanx%2BC%3F%E2%88%AB%5B3x%5E2%2F%281%2Bx%5E2%29dx%3D%E2%88%AB%5B3-3%2F%281%2Bx%5E2%29%5Ddx%3D3%E2%88%AB%5B1-1%2F%281%2Bx%5E2%29%5Ddx%3D3%5B%28x%2Bc%29-%28arctanx%2Bc%29%5D%3D3x-3arctanx%E4%B8%BA%E4%BD%95%E6%AD%A3%E7%A1%AE%E7%AD%94%E6%A1%88%E5%90%8E%E9%9D%A2%E8%BF%98%E6%9C%89%E5%8A%A0C%E5%91%A2)
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∫[3x^2/(1+x^2)dx为何是3x-3arctanx+C?∫[3x^2/(1+x^2)dx=∫[3-3/(1+x^2)]dx=3∫[1-1/(1+x^2)]dx=3[(x+c)-(arctanx+c)]=3x-3arctanx为何正确答案后面还有加C呢
∫[3x^2/(1+x^2)dx为何是3x-3arctanx+C?
∫[3x^2/(1+x^2)dx=∫[3-3/(1+x^2)]dx=3∫[1-1/(1+x^2)]dx=3[(x+c)-(arctanx+c)]=3x-3arctanx
为何正确答案后面还有加C呢
∫[3x^2/(1+x^2)dx为何是3x-3arctanx+C?∫[3x^2/(1+x^2)dx=∫[3-3/(1+x^2)]dx=3∫[1-1/(1+x^2)]dx=3[(x+c)-(arctanx+c)]=3x-3arctanx为何正确答案后面还有加C呢
因为逆积分是可以有很多个答案的.
例如 (x²+5)'=2x+0=2x.
(x²-0.124)'=2x.
两式 都是结果一样的,当逆运算时候,就需要加C.表示严谨.
∫[3x^2/(1+x^2)dx为何是3x-3arctanx+C?∫[3x^2/(1+x^2)dx=∫[3-3/(1+x^2)]dx=3∫[1-1/(1+x^2)]dx=3[(x+c)-(arctanx+c)]=3x-3arctanx为何正确答案后面还有加C呢
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