∫ 1 / X √ (x ^ 2-1) DX how to use the substitution method to calculate my two answers  

∫ 1 / X √ (x ^ 2-1) DX how to use the substitution method to calculate my two answers  

The two answers are the same, right
order
x^2 = sec^2 t = tan^2 t+1
t = arcsec x = arctan √(x^2-1)
therefore
arcsec x = arctan √(x^2-1)

Solving ∫ 1 / (1 + e∧x) DX by substitution method

1. The first kind of substitution method ∫ 1 / (1 + e ^ x) DX = ∫ e ^ (- x) / (1 + e ^ (- x)) DX = - ∫ 1 / (1 + e ^ (- x)) d (1 + e ^ (- x)) = - ln (1 + e ^ (- x)) + C = - ln ((1 + e ^ x) / e ^ x) + C = x-ln (1 + e ^ x) + C or ∫ 1 / (1 + e ^ x) DX = ∫ [1 - e ^ X / (1 + e ^ x)) DX = x - ∫ 1 / (1 + e ^ x) d (1 + e ^ x) = x -

Use the second substitution method to find the integral ∫ X / (a ^ 4 + x ^ 4) ^ (1 / 2) DX. Please tell me the process,

∫ 1 / 2 (a ^ 4 + x ^ 4) ^ (1 / 2) DX ^ 2 x ^ 2 = y ∫ 1 / 2 (b ^ 2 + y ^ 2) ^ 1 / 2) dy look up the integral table

Find the linear equation which is parallel to the line 3x + 4Y-2 = 0 and the distance from the origin is 1

From 3x + 4Y-2 = 0, y = - 3 / 4x + 1 / 2
Because it is parallel, so k = - 3 / 4
Let the intersection coordinates of the straight line and x-axis, Y-axis be a (x, 0), B (0, y) respectively, and the distance from the origin be 1, then the equations can be obtained
Xy = x & # 178; + Y & # 178; radical (1) (triangle AOB area) k = y-0 / 0-x = - 3 / 4 (2)
From (2), x = 4 / 3Y (3)
By substituting x = 4 / 3Y into (1), we can get 4 / 3Y & # 178; = 16 / 9y & # 178; + Y & # 178; radical, 4 / 3Y & # 178; = 5 / 3Y, Y1 = 5 / 4, y2 = - 5 / 4
Substituting Y1 = 5 / 4 and y2 = - 5 / 4 into (3) yields X1 = 5 / 3 and X2 = - 5 / 3
The linear equation is y = - 3 / 4x + 5 / 4 or y = - 3 / 4x-5 / 4

The function f (x) = x ^ 2 + (m-2) x + 5-m has two zeros, and they are in the interval (- 1,0) and (1,2) respectively. Find the value range of real number M
The first question is as above;
Problem 2: given that f (x) = (2 ^ x-2a) / (2 ^ x + 3a) - (1 / 3) has zero in the interval (0,1), find the value range of real number a
Problem 3: find all zeros of F (x) = (x ^ 2-5x + 6) ^ 2-7x ^ 2 + 35x-60
The above must be detailed process, otherwise do not give points, hope to help
Change the second interval of the first question to (- 3, - 2)

(1) The function f (x) = x ^ 2 + (m-2) x + 5-m has two zeros, so △ 0
Δ = B & sup2; - 4ac = (m-2) & sup2; - 4 * 1 * (5-m) > 0, the solution is m > 4 or m < - 4
Two are in the interval (- 1,0) and (1,2), so f (- 1) * f (0) < 0, f (1) * f (2) < 0
(8-2m) (5-m) ＜ 0,4 * (M + 5) ＜ 0, the solution is 4 ＜ m ＜ 5, m ＜ - 5
In order to satisfy two conditions, the intersection is taken. The value range of real number m is m ＜ - 5 or 4 ＜ m ＜ 5
（2）f(x)=(2^x-2a)/(2^x+3a)-(1/3)=2*2^x-9*a / 3*2^x+9a
F (0) * f (1) < 0
Substituting x = 0, x = 1 into f (x)
The solution is - 2 / 3 ＜ a ＜ - 1 / 3
(3) I didn't expect to go to bed

The number of intersections of parabola y = - 3x2-x + 4 and coordinate axis is ()
A. 3B. 2C. 1D. 0

The analytic formula of parabola y = - 3x2-x + 4, let x = 0, the solution is y = 4, the intersection of parabola and y-axis is (0,4), let y = 0, the solution is - 3x2-x + 4 = 0, that is 3x2 + x-4 = 0, the decomposition factor is: (3x + 4) (x-1) = 0, the solution is: X1 = - 43, X2 = 1, the intersection of parabola and x-axis is (- 43,0), (1,0), in conclusion, the number of intersection of parabola and coordinate axis is 3

It is known that the symmetry axis of parabola = AX2 + BX + C is x = 1, intersects X axis at two points a and B (a is on the left side of B), ab = 4, intersects Y axis at point C
(1) Find a point P on this parabola so that PC = Pb
(2) Find a point P on this parabola so that the triangle PBC is a right triangle with BC as a right side
(3) Find a point P on this parabola so that the triangle PBC is an isosceles triangle

The symmetric axis X = - B / (2a) = 1 = > - B / a = 2 = > b = - 2A, let y = 0, we can get ax ^ 2 + BX + C = 0x1 + x2 = - B / a = 1 / 2, x1x2 = C / AAB ^ 2 = | x1-x2 | ^ 2 = 4 ^ 2 = (x1 + x2) ^ 2-4x1x2 = (- B / a) ^ 2-4c / a = 4-4c / a = 16 = > C / a = - 3 = > C = - 3a  the parabolic equation is y = ax ^ 2-2ax-3a = a (x + 1) (x-3) intersection y

How do 16, 17, 24, 39 and 40 add up to 100

The total number of two kinds of bar is 159, because 100 + 59 = 159, so this problem is converted into several numbers, the total number is 59

If a = the square of a + the square of 5b-4ab + 2B + 100, find the minimum value of A,

A = (a-2b) square + (B + 1) square + 99
Because the first two numbers must be equal to 0, when both numbers are 0, a is the minimum
When B = - 1, a = - 2, a = 99

How to calculate X & # 178; = 5x? As long as the result

X = 0 or x = 5. That's the result