Use the first substitution method to find the indefinite integral ∫ DX / 4 + 9x ^ 2, ∫ DX / (1-2x ^ 3), ∫ a ^ Xe ^ xdx, do me a favor and write the process more specific

Use the first substitution method to find the indefinite integral ∫ DX / 4 + 9x ^ 2, ∫ DX / (1-2x ^ 3), ∫ a ^ Xe ^ xdx, do me a favor and write the process more specific

First problem, the original formula = (1 / 3) ∫ {1 / [2 ^ 2 + (3x) ^ 2]} D (3x) = 1 / 2arctan (3x / 2) + C second problem, here we need to use the multiple substitution method, which is more complex. Let me mention a point here. First, we use the cubic difference formula to expand (1-2x ^ 3), then use the rational function decomposition principle to decompose it, and then decompose the part containing x in it

The following indefinite integrals 1) ∫ √ (2 + 3x) DX 2) ∫ 4 / (1-2x) ^ 2DX 3) ∫ sin3xdx 4) ∫ DX / √ (1-25x ^ 2)
The following indefinite integrals can be obtained by the method of substitution
1）∫√(2+3x)dx
2）∫4/(1-2x)^2dx
3）∫sin3xdx
4）∫dx/√(1-25x^2)
5）∫dx/1+9x^2
6）∫cos^3xdx

1）∫√(2+3x)dxt=2+3x,x=1/3*t-2/3,dx=1/3dt）∫√(2+3x)dx=St^(1/2)*1/3dt=1/3*2/3*t^(3/2)+c=2/9*(2+3x)^(3/2)+c2）∫4/(1-2x)^2dxt=1-2x,x=-1/2*t+1/2,dx=-dt）∫4/(1-2x)^2dx=S4/t^2 *(-dt)=-4St^(-2)*dt=4/t+c...

Compare the size of 2 / a-178; - b-178; + 2 and 3 / a-178; - 2b-178; + 1
Wrong number (A & # 178; - B & # 178; + 2) / 2 and (A & # 178; - 2b & # 178; + 1) / 3

Subtraction
（a²/2-b²+2）-（a²/3-2b²+1）
=a²/6+b²+3>0
So the first big one

Molar mass of aluminum
The molar mass of aluminum is 26.98154/mol. Is it possible to have 5.0x10 ^ - 25g aluminum?

First of all, we mainly know that the atomic mass order is 10 ^ - 26kg, that is, 10 ^ - 23g, 5.0x10 ^ - 25g is below the atomic level, so there is no such Al, because its mass is not enough for an atom!

On the formula of value added tax
Sales including tax / (1 + tax rate) = sales excluding tax
What does 1 mean? Why is sales including tax divided by 1 plus tax rate equal to sales excluding tax? I have tried. It seems that the result calculated by this formula is not very consistent with that calculated directly by multiplying the value-added part by tax rate~
I'm not good at math and I don't quite understand the formula,

This formula evolved in this way
Sales including tax = sales excluding tax + VAT payable
VAT payable = sales excluding tax x tax rate
So,
Sales including tax = sales excluding tax + sales excluding tax × tax rate = sales excluding tax × (1 + tax rate)
Therefore, sales excluding tax = sales including tax / (1 + tax rate)
Here, 1 refers to sales excluding tax

For a 5V 500mA line, how much resistance should be added to reduce the voltage to 3.2 to 4V? What is the current after the step-down?

The forward voltage drop of two 1N4001 series diodes in series is about 1.4V. No matter how large the current is, the voltage drop of 1.4V is almost unchanged. You can get a voltage of 5-1.4 ≈ 3.6 (V). The maximum output current after voltage reduction is 500mA

Observe the following: 1 & sup2; + 1 = 1x2 2 & sup2; + 2 = 2x3 3 & sup2; + 3 = 3x4. The law is expressed by the formula containing natural number n (n ≥ 1)

n²+n=n(n+1)

R1 = 6 ohm, R2 = 9 ohm, R1 and R2 are connected in series in the circuit, and the voltage at both ends of resistance R2 is 9V,
1) Find the work done by the current at R1 in one minute
2) Find the work done by the current at R2 in one minute
3) Total work done by current
R1 = 6 ohm, R2 = 9 ohm, R1 and R2 are connected in parallel in the circuit, and the current is 1A through resistance R2
1) Find the work done by the current at R1 in one minute
2) Find the work done by the current at R2 in one minute
3) Total work done by current

1) I = u / r = 9 / 9 = 1a, work done by current in R1 in one minute Q1 = I * I * r * t = 1 * 1 * 6 * 60 = 360j
2) Work done by current in R2 in one minute Q2 = 1 * 1 * 9 * 60 = 540j
3)Q=360+540=900J
R1 = 6 ohm, R2 = 9 ohm, R1 and R2 are connected in parallel in the circuit, and the current at both ends of R2 is 1A through resistance;
U = IR = 1 * 9 = 9V. The current at both ends of resistance R1 is 9 / 6 = 1.5A
1) Work done by current at R1 in one minute Qi = uit = 9 * 1.5 * 60 = 810j
2) Work done by current in R2 in one minute Q2 = 9 * 1 * 60 = 540j
3)Q=810+540=1350J

How to solve the equation 8x + [6 × (13-x)] = 82
There is also an explanation, if I am satisfied, there are points!

8x+[6×(13-x)]=82
8x+78-6x=82
2x=4
x=2

Factors affecting conductor resistance
For two copper conductors with equal cross-sectional area, the ratio of length is 1:2
What is the resistance ratio?
The ratio of cross-sectional area of two copper wires of equal length is 1:2
What is the resistance ratio?

R = PL / s, which you will learn after high school... Now let's say, R is resistance, l is length, P is relative ratio, s is cross section
That is to say, the resistance is directly proportional to the relative value / length. The greater the relative value / length is, the greater the resistance is
Now understand a little bit... So for two copper wires with equal cross-sectional area, the ratio of length is 1:2, and the ratio of resistance is 1:2.. for two copper wires with equal length, the ratio of cross-sectional area is 1:2, and the ratio of resistance is 2:1