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2. 令 1/x = u, I = lim<u→0>[(3^u+5^u)/2]^(1/u)
= lim<u→0>{[1+(3^u+5^u-2)/2]^[2/(3^u+5^u-2)]}^[(3^u+5^u-2)/(2u)]
= e^lim<u→0>[(3^u+5^u-2)/(2u)] = e^lim<u→0>[(3^uln3+5^uln5)/2]
= e^[(ln3+ln5)/2] = [e^(ln15)]^(1/2) = √15
4. 定义域 x > 0 或 x < -1.
令 √[(1+x)/x] = u, 则 1+x = xu^2, x = 1/(u^2-1), dx = -2udu/(u^2-1)^2
I = ∫(u^2-1)u(-2u)du/(u^2-1)^2 = -2∫u^2du/(u^2-1)
= -2∫(u^2-1+1)du/(u^2-1) = -2∫[1+1/(u^2-1)]du = -∫[2+1/(u-1)-1/(u+1)]du
= -2u + ln|(u-1)/(u+1)| + C = -2√[(1+x)/x] + ln|{√[(1+x)/x]-1}/{√[(1+x)/x]+1}| + C
x > 0 时, I = -2√[(1+x)/x] + ln{1+2x+2√[x(1+x)]} + C
x < -1 时, I = -2√[(1+x)/x] + ln|1+2x+2√[x(1+x)]|+ C
= lim<u→0>{[1+(3^u+5^u-2)/2]^[2/(3^u+5^u-2)]}^[(3^u+5^u-2)/(2u)]
= e^lim<u→0>[(3^u+5^u-2)/(2u)] = e^lim<u→0>[(3^uln3+5^uln5)/2]
= e^[(ln3+ln5)/2] = [e^(ln15)]^(1/2) = √15
4. 定义域 x > 0 或 x < -1.
令 √[(1+x)/x] = u, 则 1+x = xu^2, x = 1/(u^2-1), dx = -2udu/(u^2-1)^2
I = ∫(u^2-1)u(-2u)du/(u^2-1)^2 = -2∫u^2du/(u^2-1)
= -2∫(u^2-1+1)du/(u^2-1) = -2∫[1+1/(u^2-1)]du = -∫[2+1/(u-1)-1/(u+1)]du
= -2u + ln|(u-1)/(u+1)| + C = -2√[(1+x)/x] + ln|{√[(1+x)/x]-1}/{√[(1+x)/x]+1}| + C
x > 0 时, I = -2√[(1+x)/x] + ln{1+2x+2√[x(1+x)]} + C
x < -1 时, I = -2√[(1+x)/x] + ln|1+2x+2√[x(1+x)]|+ C
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这个需要利用洛必达法则,当x趋近于+∞时,x趋近于+∞,e^(-x)趋近于0,变形为x/e^x
上下同时求导为1/e^x,当x趋近于+∞时,其趋近于0
上下同时求导为1/e^x,当x趋近于+∞时,其趋近于0
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能不能麻烦写哈过程
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