不定积分dx/x^2根号(4x^2-9)
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不定积分dx/x^2根号(4x^2-9)
![不定积分dx/x^2根号(4x^2-9)](/uploads/image/z/11863402-34-2.jpg?t=%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86dx%2Fx%5E2%E6%A0%B9%E5%8F%B7%EF%BC%884x%5E2-9%EF%BC%89)
令x=3/(2cosu),则:cosu=3/(2x),dx=3{sinu/[2(cosu)^2]}du.
∴∫{1/[x^2√(4x^2-9)]}dx
=∫{(2cosu/3)^2/√[9/(cosu)^2-9]}·3{sinu/[2(cosu)^2]}du
=(4/9)∫{(cosu)^2/√[9(sinu)^2/(cosu)^2]}·(3/2)[sinu/(cosu)^2]du
=(2/9)∫[(cosu)^3/sinu][sinu/(cosu)^2]du
=(2/9)∫cosudu
=(2/9)sinu+C
=(2/9)√[1-(cosu)^2]+C
=(2/9)√[1-9/(4x^2)]+C
=(2/9)√(4x^2-9)/(2x)+C
=[1/(9x)]√(4x^2-9)+C.
∴∫{1/[x^2√(4x^2-9)]}dx
=∫{(2cosu/3)^2/√[9/(cosu)^2-9]}·3{sinu/[2(cosu)^2]}du
=(4/9)∫{(cosu)^2/√[9(sinu)^2/(cosu)^2]}·(3/2)[sinu/(cosu)^2]du
=(2/9)∫[(cosu)^3/sinu][sinu/(cosu)^2]du
=(2/9)∫cosudu
=(2/9)sinu+C
=(2/9)√[1-(cosu)^2]+C
=(2/9)√[1-9/(4x^2)]+C
=(2/9)√(4x^2-9)/(2x)+C
=[1/(9x)]√(4x^2-9)+C.