不定积分计算 ∫dx/[(2x²+1)√(x²+1)]这样计算正确吗?令t=√(x²+1) 则t²=x²+1 dx=dt原式=∫dt/{[2(t²-1)+1]t}=∫dt/[(2t²-1)t]=∫[2t/(2t²-1)-1/t]dt=∫[2t/(2t²-1)]dt-∫(1/t)dt=(1/2)∫[1/(2
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![不定积分计算 ∫dx/[(2x²+1)√(x²+1)]这样计算正确吗?令t=√(x²+1) 则t²=x²+1 dx=dt原式=∫dt/{[2(t²-1)+1]t}=∫dt/[(2t²-1)t]=∫[2t/(2t²-1)-1/t]dt=∫[2t/(2t²-1)]dt-∫(1/t)dt=(1/2)∫[1/(2](/uploads/image/z/2541944-56-4.jpg?t=%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E8%AE%A1%E7%AE%97+%E2%88%ABdx%2F%5B%282x%26%23178%3B%2B1%29%E2%88%9A%28x%26%23178%3B%2B1%29%5D%E8%BF%99%E6%A0%B7%E8%AE%A1%E7%AE%97%E6%AD%A3%E7%A1%AE%E5%90%97%3F%E4%BB%A4t%3D%E2%88%9A%28x%26%23178%3B%2B1%29+%E5%88%99t%26%23178%3B%3Dx%26%23178%3B%2B1+dx%3Ddt%E5%8E%9F%E5%BC%8F%3D%E2%88%ABdt%2F%7B%5B2%28t%26%23178%3B-1%29%2B1%5Dt%7D%3D%E2%88%ABdt%2F%5B%282t%26%23178%3B-1%29t%5D%3D%E2%88%AB%5B2t%2F%282t%26%23178%3B-1%29-1%2Ft%5Ddt%3D%E2%88%AB%5B2t%2F%282t%26%23178%3B-1%29%5Ddt-%E2%88%AB%281%2Ft%29dt%3D%281%2F2%29%E2%88%AB%5B1%2F%282)
不定积分计算 ∫dx/[(2x²+1)√(x²+1)]这样计算正确吗?令t=√(x²+1) 则t²=x²+1 dx=dt原式=∫dt/{[2(t²-1)+1]t}=∫dt/[(2t²-1)t]=∫[2t/(2t²-1)-1/t]dt=∫[2t/(2t²-1)]dt-∫(1/t)dt=(1/2)∫[1/(2
不定积分计算 ∫dx/[(2x²+1)√(x²+1)]
这样计算正确吗?
令t=√(x²+1) 则t²=x²+1 dx=dt
原式=∫dt/{[2(t²-1)+1]t}
=∫dt/[(2t²-1)t]
=∫[2t/(2t²-1)-1/t]dt
=∫[2t/(2t²-1)]dt-∫(1/t)dt
=(1/2)∫[1/(2t²-1)]d(2t²-1)-∫(1/t)dt
=(1/2)ln(2t²-1)-lnt+C
再把x带回
若令x=tant则结果相差很大
上式哪一步错误?
不定积分计算 ∫dx/[(2x²+1)√(x²+1)]这样计算正确吗?令t=√(x²+1) 则t²=x²+1 dx=dt原式=∫dt/{[2(t²-1)+1]t}=∫dt/[(2t²-1)t]=∫[2t/(2t²-1)-1/t]dt=∫[2t/(2t²-1)]dt-∫(1/t)dt=(1/2)∫[1/(2
从一开始就错了
t=√(x²+1) 则t²=x²+1
x=√t^2-1
dx/dt=t/(√t^2-1)
1/[(2t²-1)t]
= A/(√2t +1) + B/(√2t -1) + C/t
dt≠dx,你直接把dt当dx解了