求怎么裂项(首先可以裂项,即∫(1/4-9x²)dx=1/4∫[1/(2+3x)+1/(2-3x)]dx=1/4×1/3×ln(3x+2)-1/4×1/3ln(3x-2)+C答案为∫(1/4-9x²)dx=1/12ln[(3x+2)/(3x-2)]+c (c为数))
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![求怎么裂项(首先可以裂项,即∫(1/4-9x²)dx=1/4∫[1/(2+3x)+1/(2-3x)]dx=1/4×1/3×ln(3x+2)-1/4×1/3ln(3x-2)+C答案为∫(1/4-9x²)dx=1/12ln[(3x+2)/(3x-2)]+c (c为数))](/uploads/image/z/3768617-65-7.jpg?t=%E6%B1%82%E6%80%8E%E4%B9%88%E8%A3%82%E9%A1%B9%EF%BC%88%E9%A6%96%E5%85%88%E5%8F%AF%E4%BB%A5%E8%A3%82%E9%A1%B9%2C%E5%8D%B3%E2%88%AB%281%2F4-9x%26%23178%3B%29dx%3D1%2F4%E2%88%AB%5B1%2F%282%2B3x%29%2B1%2F%282-3x%29%5Ddx%3D1%2F4%C3%971%2F3%C3%97ln%283x%2B2%29-1%2F4%C3%971%2F3ln%283x-2%29%2BC%E7%AD%94%E6%A1%88%E4%B8%BA%E2%88%AB%281%2F4-9x%26%23178%3B%29dx%3D1%2F12ln%5B%283x%2B2%29%2F%283x-2%29%5D%2Bc+%28c%E4%B8%BA%E6%95%B0%29%EF%BC%89)
求怎么裂项(首先可以裂项,即∫(1/4-9x²)dx=1/4∫[1/(2+3x)+1/(2-3x)]dx=1/4×1/3×ln(3x+2)-1/4×1/3ln(3x-2)+C答案为∫(1/4-9x²)dx=1/12ln[(3x+2)/(3x-2)]+c (c为数))
求怎么裂项(首先可以裂项,即∫(1/4-9x²)dx=1/4∫[1/(2+3x)+1/(2-3x)]dx=1/4×1/3×ln(3x+2)-1/4×1/3ln(3x-2)+C
答案为∫(1/4-9x²)dx=1/12ln[(3x+2)/(3x-2)]+c (c为数))
求怎么裂项(首先可以裂项,即∫(1/4-9x²)dx=1/4∫[1/(2+3x)+1/(2-3x)]dx=1/4×1/3×ln(3x+2)-1/4×1/3ln(3x-2)+C答案为∫(1/4-9x²)dx=1/12ln[(3x+2)/(3x-2)]+c (c为数))
∫(1/4-9x²)dx=1/4∫[1/(2+3x)+1/(2-3x)]dx=1/4×1/3×ln(3x+2)-1/4×1/3ln(3x-2)+C
由题意得:这是分式加法公式的逆应用
(1)(4-9x^2)=(2-3x)(2+3x)
(2)设1/(4-9x^2)=a/(2-3x)+b/(2+3x)对任意的有意义的x恒成立
即而a/(2-3x)+b/(2+3x)=[(2a+2b)+(6a-6b)x]/(4-9x^2)
所以(2a+2b)=1 且6a-6b=0
解得 a=b=1/4
即1/(4-9x^2)=1/4[1/(2-3x)+1/(2+3x)]
1/(4-9x^2)=1/(2+3x)(2-3x)=1/4[1/(2-3x)+1/(2+3x)]
所以有:∫(1/4-9x²)dx=1/4∫[1/(2+3x)+1/(2-3x)]dx=1/4×1/3×ln|(3x+2)|-1/4×1/3ln|(3x-2)|+C=1/12ln|(3x+2)/(3x-2)|+C
1/(4-9x^2)=1/(2+3x)(2-3x)=1/4[1/(2-3x)+1/(2+3x)]
所以有:∫(1/4-9x²)dx=1/4∫[1/(2+3x)+1/(2-3x)]dx=1/4×1/3×ln|(3x+2)|-1/4×1/3ln|(3x-2)|+C=1/12ln|(3x+2)/(3x-2)|+C