The sequence xn = (1 + 2) ^ (- 1) + (1 + 2 ^ 2) ^ (- 1) + (1 + 2 ^ 3) ^ (- 1) +... + (1 + 2 ^ n) ^ (- 1) proves that {xn} converges

The sequence xn = (1 + 2) ^ (- 1) + (1 + 2 ^ 2) ^ (- 1) + (1 + 2 ^ 3) ^ (- 1) +... + (1 + 2 ^ n) ^ (- 1) proves that {xn} converges

xn=(1+2)^(-1)+(1+2^2)^(-1)+(1+2^3)^(-1)+...+(1+2^n)^(-1)

The second kind of indefinite integral substitution integral method inverse substitution ∫ DX / X (x ^ 7 + 2) Let x = 1 / T, DX = - DT / T ^ 2 ∫dx/x（x^7+2） =∫t/(1/t)^7+2*(-1/t^2)dt =-∫(t^6/1+2t^7)dt =-1/14ln|1+2t^7|+c =-1/14ln|2+x^7|+1/2ln|x|+c How did you make the transition between the last two steps? I didn't understand it for a long time

(- 1/14)ln|1 + 2t⁷| + C
= (- 1/14)ln|1 + 2(1/x⁷)| + C
= (- 1/14)ln|(x⁷ + 2)/x⁷| + C
= (- 1/14)[ln|x⁷ + 2| - ln|x⁷|] + C
= (- 1/14)ln|2 + x⁷| - (- 1/14)(7ln|x|) + C
= (- 1/14)ln|2 + x⁷| + (1/2)ln|x| + C

How to find the indefinite integral of DX / x square + X + 1 by triangular substitution

∫dx/(x^2+x+1)
=∫dx/[(x+1/2)^2+3/4]
=1/(√3/2)*artan[(x+1/2)/(√3/2)]+C

On variable substitution Y = √ (x Λ 2 + a Λ 2), a is greater than 0, can x = atant, t belong to - 1 to 1? Or just let t = + 1 or - 1? It seems that x can get all real numbers.

But this kind of question ∫ (the upper limit is positive infinity, and the lower limit is 1)) [1 / X √ (x-1). This is because if you use √ (x ^ 2-1) = t, and finally get x = √ (T ^ 2 + 1), it will be the same as above

Ordinary differential equations. Variable substitution problems dy/dx=（2x^3+3x*y^2+x)/(3x^2*y+2y^3-y) But it's a sign.. it's not a total differential equation... (and I just learned ordinary differential)... I haven't learned the solution of total differential yet

（2x^3+3x*y^2+x)dx+[-(3x^2*y+2y^3-y)]dy=0
Look at page 283 of the fifth edition of the book of advanced mathematics, and the formula comes out
It's too hard
If P (x, y) DX + Q (x, y) dy = du (x, y), PDX + Qdy = 0 is called a total differential equation. Obviously, the general solution of the equation is u (x, y) = C (C is an arbitrary constant)
According to the total differential quadrature theorem of binary functions: let the open region G be a simply connected domain and the functions P (x, y) and Q (x, y) have a first-order continuous partial derivative in G, then p (x, y) DX + Q (x, y) dy is the total differential of a function U (x, y) in G if and only if p '(y) = q' (x), which is constant in G
Example: judge whether equation (3x26xy2) DX + (4y3 + 6x2y) dy = 0 is a total differential equation and find its general solution
（3x^2+6xy^2)dx+(4y^3+6x^2y)dy=0,
P=3x^2+6xy^2,Q=4y^3+6x^2y,
δ P/ δ y=12xy= δ Q/ δ x,
So this is a total differential equation,
u(x,y)=∫[0,x](3x^2+6xy^2)dx+∫[0,y]4y^3dy
=x^3+3x^2y^2+y^4,
x^3+3x^2y^2+y^4=C.

High school function variable substitution method Given that x ≠ 0, the function f (x) satisfies f (x-1 / x) = x ^ 2 + 1 / x ^ 2, then the expression of F (x) is () Analysis: if X-1 / x = t, then x ^ 2 + 1 / x ^ 2 = T ^ 2 + 2 ∴f(t)=t^2+2,∴f(x)=x^2+2 1. How x is expressed by the formula containing T needs to be calculated 2. F (T) = T ^ 2 + 2, f (x) = x ^ 2 + 2, why can t and X replace each other?

1. Personally, I think it's better to use the collocation method
x ²+ 1/x ²= (x－1/x) ²＋ 2, so f (x) = X ²＋ 2；
2. For example, f (x) = 2x-3 and f (T) = 2t-3 represent the same function
Note: for the same function, the key is that the three elements are the same, i.e. definition field, value field and corresponding law. These two functions are the same function and have nothing to do with the variable letter you use. Generally, it is customary to use X as the independent variable