求不定积分 ∫5/(x²+4x+9)dx
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求不定积分 ∫5/(x²+4x+9)dx
![求不定积分 ∫5/(x²+4x+9)dx](/uploads/image/z/6192947-11-7.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86+%E2%88%AB5%2F%28x%26%23178%3B%2B4x%2B9%29dx)
x^2+4x+9 = (x+2)^2+5
let
x+2 = √5tana
dx =√5(seca)^2 da
∫5/(x²+4x+9)dx
=∫[5/(5(seca)^2)] √5(seca)^2 da
=√5a+ C
=√5arctan[(x+2)/√5] + C
let
x+2 = √5tana
dx =√5(seca)^2 da
∫5/(x²+4x+9)dx
=∫[5/(5(seca)^2)] √5(seca)^2 da
=√5a+ C
=√5arctan[(x+2)/√5] + C