∫[√(x+3/x-1)-√(x-1/x+3)]dx求大神帮我解答下我的疑问∫[√(x+3/x-1)-√(x-1/x+3)]dx用换元法u=√((x+3)/(x-1))算到“∫[8/(u^2-1)]du”这步,下面为什么我用凑微分法 凑得 ∫[8/(u^2-1)]du=∫[4/(u^2-1)]d(u^2-1) 然
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![∫[√(x+3/x-1)-√(x-1/x+3)]dx求大神帮我解答下我的疑问∫[√(x+3/x-1)-√(x-1/x+3)]dx用换元法u=√((x+3)/(x-1))算到“∫[8/(u^2-1)]du”这步,下面为什么我用凑微分法 凑得 ∫[8/(u^2-1)]du=∫[4/(u^2-1)]d(u^2-1) 然](/uploads/image/z/1235978-26-8.jpg?t=%E2%88%AB%5B%E2%88%9A%28x%2B3%2Fx-1%29-%E2%88%9A%28x-1%2Fx%2B3%29%5Ddx%E6%B1%82%E5%A4%A7%E7%A5%9E%E5%B8%AE%E6%88%91%E8%A7%A3%E7%AD%94%E4%B8%8B%E6%88%91%E7%9A%84%E7%96%91%E9%97%AE%E2%88%AB%5B%E2%88%9A%28x%2B3%2Fx-1%29-%E2%88%9A%28x-1%2Fx%2B3%29%5Ddx%E7%94%A8%E6%8D%A2%E5%85%83%E6%B3%95u%3D%E2%88%9A%28%28x%2B3%29%2F%28x-1%29%29%E7%AE%97%E5%88%B0%E2%80%9C%E2%88%AB%5B8%2F%28u%5E2-1%29%5Ddu%E2%80%9D%E8%BF%99%E6%AD%A5%2C%E4%B8%8B%E9%9D%A2%E4%B8%BA%E4%BB%80%E4%B9%88%E6%88%91%E7%94%A8%E5%87%91%E5%BE%AE%E5%88%86%E6%B3%95+%E5%87%91%E5%BE%97+%E2%88%AB%5B8%2F%28u%5E2-1%29%5Ddu%3D%E2%88%AB%5B4%2F%28u%5E2-1%29%5Dd%28u%5E2-1%29+%E7%84%B6)
∫[√(x+3/x-1)-√(x-1/x+3)]dx求大神帮我解答下我的疑问∫[√(x+3/x-1)-√(x-1/x+3)]dx用换元法u=√((x+3)/(x-1))算到“∫[8/(u^2-1)]du”这步,下面为什么我用凑微分法 凑得 ∫[8/(u^2-1)]du=∫[4/(u^2-1)]d(u^2-1) 然
∫[√(x+3/x-1)-√(x-1/x+3)]dx求大神帮我解答下我的疑问
∫[√(x+3/x-1)-√(x-1/x+3)]dx用换元法u=√((x+3)/(x-1))算到“∫[8/(u^2-1)]du”这步,下面为什么我用凑微分法 凑得 ∫[8/(u^2-1)]du=∫[4/(u^2-1)]d(u^2-1) 然后直接得到 4ln|(u^2-1)|+C 就算化简 得4ln[(u-1)(u+1)]+C也和答案4ln[(u-1)/(u+1)]+C 差了个符号
∫[√(x+3/x-1)-√(x-1/x+3)]dx求大神帮我解答下我的疑问∫[√(x+3/x-1)-√(x-1/x+3)]dx用换元法u=√((x+3)/(x-1))算到“∫[8/(u^2-1)]du”这步,下面为什么我用凑微分法 凑得 ∫[8/(u^2-1)]du=∫[4/(u^2-1)]d(u^2-1) 然
∫[8/(u^2-1)]du = ∫[4/(u^2-1)]d(u^2-1)
这一步错了
d(u^2 -1) = 2u du
右边 = ∫ [ 8u / (u^2-1) ] du
不等于左边
正确做法:
∫ [ 8 / (u^2-1) ] du
= ∫ 4* [ 1/(u-1) - 1/(u+1) ] du
= 4 [ ln(u-1) - ln(u+1) ] + C
= 4 ln [ (u-1) / (u+1) ] + C
你是说的结果中的绝对值符号么,如果是要注意u是有范围的,至少应该是u大于1