Finding definite integral 1, ∫ (1-2) {[(INX) ^ 2] / (x ^ 3)} DX 2, ∫ (0-1) {X / [e ^ (5x)]} DX integration by parts

Finding definite integral 1, ∫ (1-2) {[(INX) ^ 2] / (x ^ 3)} DX 2, ∫ (0-1) {X / [e ^ (5x)]} DX integration by parts

1.∫ { [(Inx)^2]/(x^3) } dx = (-1/2) ∫ (Inx)^2 d x^(-2)
= (-1/2) [ (Inx)^2 * x^(-2) ] + ∫ 2Inx * x^(-3) dx
= (-1/2) [ (Inx)^2 * x^(-2) ] - ∫ Inx d x^(-2)
= (-1/2) [ (Inx)^2 * x^(-2) ] - lnx ^ x^(-2) + ∫ x^(-3) dx
= (-1/2) [ (Inx)^2 * x^(-2) ] - lnx ^ x^(-2) - (1/2) x^(-2) + C
Original formula = (- 1 / 8) (LN2) ^ 2 - LN2 / 4 - 1 / 8
2.∫ x * e^(-5x) dx = (-1/5) ∫ x d e^(-5x) = (-1/5) x * e^(-5x) + (1/5) ∫ e^(-5x) dx
= (-1/5) x * e^(-5x) + (1/25) e^(-5x) + C
The original formula = (- 4 / 25) e ^ (- 5) - 1 / 25

If the area of a rectangle is 3a2-3ab + 6a and one side is 3a, its perimeter is______ ．

Its perimeter is: 2 [(3a2-3ab + 6a) △ 3A + 3A] = 2 (a-b + 2 + 3a), = 2 (4a-b + 2), = 8a-2b + 4

Junior high school physics problems about electric power and electric energy
Suppose that when a certain lightning occurs, the voltage formed is 3.6 times 10 to the 7th power V, the current passing through the air in 0.001s is 10 to the 4th power a, the discharge power is kW, and the discharge energy is kW / h
How to calculate?
Discharge power: how many kW, discharge energy: how many kW / h

P=UI=3.6*10^7*10^4=3.6*10^11W=3.6*10^8KW
w=Pt=3.6*10^11*0.001=3.6*10^8J=100kW/h

5X+6Y=500,4X+1Y=5Y+1X,X＞Y,X=?,Y=?

5x 6y = 500 multiply both sides of the equation by 3 to get 15x 18y = 1500
4X 1y = 5Y 1x deduces 3x-4y = 0 and multiplies both sides of the equation by 5 to get 15x-20y = 0
So 15x 18y-15x 20Y = 1500
Y＝125/3 X＝500/9

In junior high school physics, over work = current × voltage × power on time, i.e. w = uit. Similarly, electric power P = UI. But u and I are both physical quantities of electricity. Why multiply them

Power refers to the rate of doing work, that is, the output / input energy per unit time. No matter you push a box or the current passes through a resistance, there is energy output, so there is the concept of power. The voltage unit "volt" is originally defined by dividing the power unit by the current unit: in a uniform conductor with constant temperature and width

What is the elastic potential energy stored in k1k2
The upper end of the light-weight spring with stiffness coefficient K2 is bolted to block 2, and the lower end is pressed on the table. The whole system is in balance. Try to calculate the elastic potential energy stored in springs K1 and K2 at this time?
The answer is 1 stored potential energy: 1 / 2 * K1 * X1 ^ 2 + (M1 ^ 2G ^ 2 / 2K1)
2:1/2*k2*x2^2+(m1+m2)^2*g^2/2k2
One: I don't understand why it is like this. I don't even know how to think about it
Second, I don't understand why the potential energy of K2 is (M1 + m2) g, not the gravitational potential energy storage
Please help us to answer these two questions. I'm satisfied with them
http://zhidao.baidu.com/question/66595689.html
I'm sorry I didn't type the title. I'm totally sorry~~~

The force analysis of the above object M1 shows that the system is balanced, the resultant force is zero, the gravity and the elastic force are equal, and the direction is opposite. So m1g = k1x1, where X1 is the amount of compression. In this way, we know that the elastic potential energy is 1 / 2 * K1 * X1 ^ 2 = (M1 ^ 2G ^ 2 / 2K1)
In the same way, M1 and M2 and spring 1 are regarded as a system and placed on spring 2. The force is balanced. The gravity is (M1 + m2) g and the elastic force is k2x2, so (M1 + m2) g = k2x2, where X2 is the amount of compression. In this way, we know that the elastic potential energy is 1 / 2 * K2 * x2 ^ 2 = (M1 + m2) ^ 2 * G ^ 2 / 2k2

What is resistance, capacitance and inductance
Easy. Thank you

Resistance definition: the resistance of a conductor is the resistance of the conductor. Resistance is the most commonly used component in all electronic circuits. The main physical feature of resistance is to change electrical energy into heat energy. It can also be said that it is an energy consuming component. The current will generate internal energy through it. Resistance usually acts as a voltage divider in the circuit

The mass of the wood block is 1kg, the stiffness coefficient of the light spring is 100N / m, the upper wood block is pressed on the upper spring, and the whole system is in a state of balance. The compression of the spring G is taken as 10

1kg=10N
So the compression is 10 △ 100 = 0.1M

In the right angle △ ABC, the sum of the two right angles is 14, and the area of the right triangle is calculated

Let BC length be a and AC length be B
From the meaning of the title
a+b=14
a^2+b^2=(2x)^2
Triangle area s = 1 / 2 * a * B
Just solve the equation~·

A spring scale is placed on the horizontal ground, q is connected with the upper end of the light spring, P is a weight, the known mass of P is 10.5kg, the mass of Q is
A spring scale is placed on the horizontal ground, q is connected with the upper end of the light spring. P is a heavy weight. It is known that the mass of P is 10.5kg, the mass of Q is 1.5kg, the mass of the spring is not included, and the stiffness coefficient is 800N / m. The system is at rest. Now, a vertical upward force is applied to p to make it do uniform acceleration from the rest. It is known that in the first 0.2S time, f is a variable force, and after 0.2S, f is a constant force, Finding the minimum and maximum of F force

The static spring is compressed, x = mg / k = (10.5 + 1.5) * 10 / 800 = 0.15m
First 0.2S
a=2S/t^2=2*0.15/0.2^2=7.5m/s2
F+F1-mg=ma,
F1max, Fmin = ma = 78.75n
After 0.2S, f is a constant force, and the spring has returned to its original length, f-mg = ma, Fmax = 183.75n