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求不定积分1/（9x^2-4)^1/2
题目详情
求不定积分 1/（9x^2-4)^1/2
▼优质解答
答案和解析
∫ 1/√(9x² - 4) dx
令x = (2/3)secy,dx = (2/3)secytany dy
原式= ∫ 1/(2tany) * (2/3)secytany dy
= (1/3)∫ secy dy
= (1/3)ln|secy + tany| + C
= (1/3)ln|(3x/2) + √(9x²-4)/2| + C
= (1/3)ln|3x+√(9x²-4)| + C
令x = (2/3)secy,dx = (2/3)secytany dy
原式= ∫ 1/(2tany) * (2/3)secytany dy
= (1/3)∫ secy dy
= (1/3)ln|secy + tany| + C
= (1/3)ln|(3x/2) + √(9x²-4)/2| + C
= (1/3)ln|3x+√(9x²-4)| + C
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