The problem of integral variable substitution of univariate function? If it is such a question ∫ (the upper limit is positive infinity, the lower limit is 1) [1 / X √ (x-1)] DX, you can use √ (x-1) = t as a variable substitution But this question ∫ (the upper limit is positive infinity, the lower limit is 1)) [1 / X √ (x-1)] DX =? It is necessary to use x = 1 / T, but not √ (x square - 1) = t as in the above question, Why?

The problem of integral variable substitution of univariate function? If it is such a question ∫ (the upper limit is positive infinity, the lower limit is 1) [1 / X √ (x-1)] DX, you can use √ (x-1) = t as a variable substitution But this question ∫ (the upper limit is positive infinity, the lower limit is 1)) [1 / X √ (x-1)] DX =? It is necessary to use x = 1 / T, but not √ (x square - 1) = t as in the above question, Why?

This is because if you use √ (x ^ 2-1) = t, and finally get x = √ (T ^ 2 + 1), you will not be able to integrate as above

∮dx/(√a ²+ x ²) This is a calculus formula problem, ∮ is the integral sign =∫sec θ d θ =ln|sec θ+ tan θ|+ C,tan θ= X / A, bevel = √ (a) ²+ x ²) How does the previous step translate to the next step?

∫ 1/√(a ²+ x ²) DX, a are arbitrary constants
Let x = atan θ, dx=asec ²θ d θ
=a∫sec ²θ/ √(a ²+ a ² tan ²θ) d θ
=a∫sec ²θ/ √[a ² (1+tan ²θ)] d θ
=a∫sec ²θ/ √(a ² sec ²θ) d θ
=a∫sec ²θ/ (asec θ) d θ
=∫sec θ d θ
=ln|sec θ+ tan θ|+ C,tan θ= X / A, bevel = √ (a) ²+ x ²)
=ln|[√(a ²+ x ²)/ a+x/a|+C
=ln|[x+√(a ²+ x ²)]/ a|+C
=ln|2x+2√(a ²+ x ²)|+ C

Calculus of advanced mathematics Given that a is a constant real number and Y is a function of X, find the solution of the following differential equation, y '' (x) + 2ay '(x) + y (x) = 2, which satisfies y (0) = 0 and y' (0) = 2A. Here y 'and y' 'are the first and second derivatives of X respectively

y''+2ay'+y=2
y''+2ay'+(y-2)=0 (y-2)''=y'' (y-2)'=y'
(y-2)''+2a(y-2)'+(y-2)=0
characteristic equation
r^2+2ar+1=0
(r+a)^2=a^2-1
r1=-a+√(a^2-1) r2=-a-√(a^2-1)
general solution
y-2=C1e^[-a+√(a^2-1)]x +C2e^[-a-√(a^2-1)]x
y(0)=0
C1+C2+2=0
y'(0)=C1[-a+√(a^2-1)]+C2[-a-√(a^2-1)]=2a
C1=-1
C2=-1
Special solution y = (- 1) e ^ [- A + √ (a ^ 2-1)] x + (- 1) [1 / √ (a ^ 2-1) - 1] e ^ [- A - √ (a ^ 2-1)] x + 2

Solve mathematical calculus y''=4x-y'

y''+y'=4x
First solve the homogeneous equation
y''+y'=0
Let z = y '
z'+z=0
z=Ae^(-x)
y'=Ae^(-x)
integral
y=Ae^(-x)+B
Re solution nonhomogeneous
y''+y'=4x
y=ax^2+bx+c
2a+2ax+b=4x
2a=4,2a+b=0
a=2,b=-4
So the special solution is 2x ^ 2-4x
So the solution y = AE ^ (- x) + B + 2x ^ 2-4x

The following are basic integral problems:   An object is moving with velocity (in FT / sec) V (T) = T ^ 2-3t-10 Find the displacement and total distance travelled from t=0 to t=8. Find the displacement and distance Evaluate the infinite integral Estimate the area under the graph of f(x)=1/(x+2) over the interval [0,5] ,using four approximating rectangles and right endpoints. Sum RN   Repeat the approximation using left endpoints-----Ln. Find the integral of 4x (X-2) ^ (1 / 2)

(1) Displacement is the integral of velocity. Note that these two are vectors, which can be calculated according to the appropriate rules. If the initial velocity is - 10 and the initial displacement is 0, the unit will be ignored. The indefinite integral of V to t is 2 / 3 · T ^ 3-3 / 2 · T ^ 2-10t + S. the left formula is the expression of displacement, which directly brings in two T, and the difference is the displacement. As for the distance, it is equivalent to the distance, Then, it is the sum of the displacements of the two segments moving in opposite directions. Make it 0 through the velocity expression, and the zero velocity time is the fifth second. Therefore, refer to the displacement method to add the two motion distances
(4) For the common substitution method, after reduction, make the formula look better, so that t = (X-2) ^ (1 / 2), t is greater than or equal to 0, and the original formula is 4 (T ^ 2 + 2) · T. integrate it to get a formula, and I won't know the next steps

First question ∫ f (AX + b) DX = 1 / a ∫ f (U) Du (a ≠ 0, u = ax + b). How is this result calculated? Second question Let ∫ f (x) DX = insinx + C, find ∫ XF (1-x ^ 2) DX

This is the first commutative integration, let u = ax + B, Du = ADX, DX = 1 / Adu
∫f(ax+b)dx=1/a∫f(u)du
2) Let u = 1-x ^ 2, Du = - 2xdx, XF (1-x ^ 2) DX = - 1 / 2F (U) Du
∫xf(1-x^2)dx=1/2∫f(1-x^2)d(x^2)=-1/2∫f(1-x^2)d(1-x^2)=(-1/2)lnsin(1-x^2)+C