LNX reciprocal integral How to find the integral of 1 / LNX? There are also limits for the sum of sequences [(k Power of a) / k], where a = 2 / 2 root sign 2, k = 1,2,3,. N, n tend to infinity

LNX reciprocal integral How to find the integral of 1 / LNX? There are also limits for the sum of sequences [(k Power of a) / k], where a = 2 / 2 root sign 2, k = 1,2,3,. N, n tend to infinity

How to find the integral of 1 / LNX
There is only a series solution
It's a transcendental function. There's no elementary representation
There's a special logarithmic integral for this function
Finding the limit of sequence sum [(k Power of a) / k],
lim(a+a^2/2+a^3/3+...+a^k/k)=-ln(1-√2/2)

General solution of differential equation y '' '= LNX

Let y = x / LNX be the solution of the differential equation y '= Y / x + φ (x / y), then the expression of φ (x / y) is?
Take y = x / LNX into the equation y '= Y / x + φ (x / y) to get: 1 / lnx-1 / (LNX) ^ 2 = 1 / LNX + φ (LNX) to get: φ (LNX) = - 1 / (LNX) ^ 2, then φ (x / y) = - y ^ 2 / x ^ 2
I think he first changed the band of φ (x / y) to that of φ (LNX), and then set it to that of φ (x / y). Did he make a detour?
Why not calculate φ (x / y) directly? For example, if 1 / lnx-1 / (LNX) ^ 2 = 1 / LNX + φ (x / y), then φ (x / y) = - 1 / (LNX) ^ 2
I know it's not right, but I don't know why it's not right. Why do we have to row around and do unnecessary things?

Because when you substitute y, the part related to y in the expression of φ (x / y) is also replaced, so - 1 / (LNX) ^ 2 actually has the part of X and Y (x / LNX) at the same time. You need to substitute y in order to find out the relationship

X approaches infinity LIM (1-3 / x) ^ x=

With exponential function, the base is e, the index is x * ln (1-3 / x), and because the equivalent infinitesimal ln (1-3 / x) = (- 3 / x), this is a formula. You should know that, then the base is e, and the index is x * (- 3 / x) = - 3, so the final result is e ^ (- 3)

Calculation: (1 / 3-1 / 2) / 1 and 1 / 4 / 1 / 10

(1 / 3-1 / 2) / 1 and 1 / 4 / 1 / 10
=-1 / 6 × 5 / 4 × 10
=-5 / 24 × 10
=-25 out of 12

Find LIM (1 / x + 2 ^ 1 / x) ^ x to approach infinity find Lim {[a ^ (1 / N) + B ^ (1 / N)] / 2} ^ n a, b greater than zero, X to approach infinity

The second Lim {[a ^ (1 / N) + B ^ (1 / N)] / 2} ^ n
=e^lim n*ln{[a^(1/n)+b^(1/n)]/2}
And LN {[a ^ (1 / N) + B ^ (1 / N)] / 2} = ln {1 + {[a ^ (1 / N) + B ^ (1 / N)] / 2} - 1} ~ {a ^ (1 / N) + B ^ (1 / N)] / 2} - 1}
The original formula is e ^ LIM (n / 2) * [a ^ (1 / N) - 1 + B ^ (1 / N) - 1]
And a ^ X-1 = xlna + O (x)
So ^ / LNA = (1 / LNA) + (1)
The original formula is changed to e ^ LIM (n / 2) * [a ^ (1 / N) - 1 + B ^ (1 / N) - 1]
=e^lim(n/2)*[(1/n)*(lnab)+o(1/n)]
=e^lim ln(ab)^(1/2)
=(ab)^(1/2)

Calculation: (1-1 / 2) 2 (1-1 / 3) 2 (1-1 / 4) 2 × ×（1-1/10）2
Next to each bracket is the square! （1-1/2）^2（1-1/3）^2（1-1/4）^2×…… ×（1-1/10）^2
What you wrote is so messy. I didn't write it wrong

(1/2×2/3×3/4...×9/10)²=1/100

What is the indefinite integral of ∫ [arc (TaNx * TaNx) / (1 + X * x)] DX?

Is this a wrong question? It should be like this
∫[（arc tanx ）^2/ (1+x*x) ] dx
=∫(arc tanx )^2d(arctanx)
=arctanx^3/3+C

Calculation: (1-1 / 2) * (1-1 / 3) * (1-1 / 4) * *(1-1/10).

=(1/2)*(2/3)*(3/4)*…… *(9/10)
The first denominator and the next numerator are approximately divided
=1/10
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The sum of two reciprocal numbers equals five sixths. The difference between the two numbers is one. What are the two numbers?

n+m=5/6,1/n-1/m=1
n=3,m=2/5