Seeking indefinite integral ∫ 1 / 1 + 4x ^ 2 DX

Seeking indefinite integral ∫ 1 / 1 + 4x ^ 2 DX

The original formula = 1 / 2 * ∫ 1 / [1 + (2x) & sup2;] d (2x)
=(1/2)arctan(2x)+C

Seeking indefinite integral ∫ ((5-4x) ∧ 3) DX

∫ (5 - 4x)^3 dx
= (1/4)∫ (5 - 4x)^3 d(4x)
= (- 1/4)∫ (5 - 4x)^3 d(- 4x)
= (- 1/4)∫ (5 - 4x)^3 d(5 - 4x)
= (- 1/4) * (5 - 4x)^(3 + 1)/(3 + 1) + C
= (- 1/16)(5 - 4x)^4 + C
Make up the differential step by step

As shown in the figure, the triangle ABC is inscribed in the circle O, ab = AC, and the chord ad intersects BC at the point E, AE = 4, ed = 5.1. Find AC 2. If I is a point on ad, AI = AC, prove:
I is the heart of △ BCD

Do you have a picture

If f (x) = 2x ^ 3-3x ^ 2 + ax + B divided by X + 1, the remainder is 7 and divided by X-1, the remainder is 5, try to find the value of a and B

f(x)=2x^3-3x^2+ax+b
=2x^3+2x^2-5x^2+ax+b
=2x^2*(x+1)-5x^2-5x+5x+ax+b
=2x^2*(x+1)-5x(x+1)+(a+5)x+(a+5)-a-5+b
=2x^2*(x+1)-5x(x+1)+(a+5)(x+1)+(b-a-5)
So: b-a-5 = 7 ------ (1)
f(x)=2x^3-3x^2+ax+b
=2x^3-2x^2-x^2+ax+b
=2x^2*(x-1)-x^2+x+(a-1)x+b
=2x^2*(x-1)-x(x-1)+(a-1)(x-1)+(b+a-1)
So: a + B-1 = 5 ------ (2)
Simultaneous (1) and (2) are as follows:
a=-3,b=9

A is the smallest positive integer, B is the opposite number of the largest negative integer, C is the number with the smallest distance on the number axis, find the value of a + 2B + C

A=1 B=1 C=0
A+2B+C=3

Merging congeners: 3A's Square b-5a's square B

3a²b -5a²b =-2a²b

380V electricity without zero line, how can the ground wire become 220V

There are two ways
1. Take the earth as the zero line directly, and the grounding shall be good. If any one of the three phases is selected, the voltage of 220 V will be obtained
2. Buy three bulbs with socket or other loads of the same power, connect the three bulbs according to the star type, and the voltage between the neutral point and the phase line is 220 v

Write more. 30 is about the same

You should go to Baidu Library to find, arbitrary download

A capacitor is connected in series with a resistor connected to 220 V AC. what is the power in this path? The resistor is 39 and the capacitor is 20 UF

Capacitive reactance:
Xc＝1/2πfC＝1/(2×3.14×50×0.00002)≈159(Ω)
Series circuit impedance:
Z = root sign (r square + XC Square) = root sign (39 × 39 + 159 × 159) ≈ 164 (Ω)
Circuit current:
I＝U/Z＝220/164≈1.34(A)
Power:
P＝I×I×R＝1.34×1.34×39≈70(W)

Divide the following fractions
x/xy-y，y/xy+y

x/xy-y=(x-xy)²/xy=x(1-y²)/xy=(1-y²)/xy
y/xy+y=(y+xy²)/xy=y(1+xy)/xy=(1+xy)/xy