How to calculate acceleration with odd displacement by successive difference method

How to calculate acceleration with odd displacement by successive difference method

If there are odd displacement segments, one segment at both ends or one segment in the middle can be omitted. If the displacement increases from small to large, there are S1, S2, S3, S4, S5 in turn. The time of each displacement segment is t, S1 or S3 or S5 can be omitted. If S1 is omitted, then a = Δ s / T ^ 2 = {[(s4-s2) + (s5-s3)}

Why is the displacement of a small ball falling from 5 meters above the ground 5 meters, not - 5 meters

In this case, the scene generally defaults to the positive direction downward

How to calculate the combined acceleration
"The acceleration of an object moving separately on the x-axis is 1 m / S2, and the acceleration of moving separately on the y-axis is 0, so the combined acceleration of an object is a = 1 m / S2". How can we get the "combined acceleration" here? How can we get the combined acceleration? What formula can we use?

Using the method of vector summation, we can solve the problem in a parallelogram

Indefinite integral of Xe ^ (x ^ 2)

∫ x*e^(x^2) dx = 1/2* ∫ e^(x^2) d(x^2) = 1/2*e^(x^2)+C

Calculate indefinite integral ∫ DX / &; (x) + √ (x))

Let a = x ^ (1 / 6)
Then x = a ^ 6
dx=6a^5da
Then the denominator = A & sup2; + A & sup3;
The original formula = ∫ 6A ^ 5da / (A & sup2; + A & sup3;)
=6∫a³da/(a+1）
=6∫(a³+1-1)da/(a+1）
=6∫[(a+1)(a²-a+1)-1]da/(a+1）
=6∫[(a²-a+1-1/(a+1]da
==6[a³/3-a²/2+a-ln(a+1)]+C
=2√x-3³√x+x^(1/6)-ln[x^(1/6)+1]+C

Seeking indefinite integral ∫ ((1 + 2lnx) / x) DX

∫((1+2lnx)/x)dx
=∫((1+2lnx)dlnx
=1/2∫(1+2lnx)d(1+2lnx)
=(1+2lnx)²/4+C

∫1/{x(1+2lnx)}dx

∫1/{x(1+2lnx)}dx
=∫1/(1+2lnx)dlnx
=1/2*∫1/(1+2lnx)d(1+2lnx)
=1/2*ln(1+2lnx)+c

If the function f (x) = e ^ 2x, then the indefinite integral f (x / 2) DX =?

∫f(x/2)dx=e^x+C

"Let f (x) = X2, find the indefinite integral f '(2x) DX =. The indefinite integral f (2x) DX ='"
Sign for derivation

(1) The quadratic power of 4 times x
(2) Four Thirds of X to the third power,

Find the indefinite integral of (TaNx) ^ 5! To process! Absolutely give points

∫ (tanx)^5 dx=∫ tan³xtan²x dx=∫ tan³x(sec²x-1) dx=∫ tan³xsec²x dx - ∫ tan³x dx=∫ tan³x d(tanx) - ∫ tanx(sec²x-1) dx=(1/4)(tanx)^4 - ∫ tanxsec²...