How to calculate indefinite integral for 1 / (x (x ^ 7 + 2) with the second substitution method? (let x = 1 / T)

How to calculate indefinite integral for 1 / (x (x ^ 7 + 2) with the second substitution method? (let x = 1 / T)

First, we use the second kind of substitution method, and then we use the differential method
x=1/t dx=-1/t²dt
The original formula = ∫ 1 / [(1 / T) * (1 / T ^ 7 + 1] * (- 1 / T & # 178;) DT = - ∫ (T ^ 6) / (1 + T ^ 7) DT = (- 1 / 7) J ∫ 1 / (1 + T ^ 7) d (1 + T ^ 7)
-(1/7)ln(1+t^7) +C
=-(1/7)ln[1+(1/x)^7] +C
=-(1/7)ln(1+x^7) +lnx +C

The second substitution method is used to solve the indefinite integral of X / 1 + x ^ 4

Let X & # 178; = Tan θ, 2x DX = sec & # 178; θ D θ∫ X / (1 + x ^ 4) DX = ∫ X / (1 + Tan & # 178; θ) &; sec & # 178; θ D θ / (2x) = (1 / 2) ∫ sec & # 178; θ / sec & # 178; θ D θ = (1 / 2) θ + C = (1 / 2) arctan (X & # 178;) + C

How to calculate the integral of ∫ cos X / x) DX
How to calculate the integral of ∫ (COS X / x) DX

It is not an elementary function
If it is not required, it can be expanded into an infinite series and then integrated term by term

1/(1+cos x) dx
The question mark doesn't show. It's the integral symbol

∫ 1/(1+cos x) dx
=∫ (1-cosx)/[ 1-(cosx)^2 ] dx
=∫ [ 1/(sinx)^2 - cosx/(sinx)^2 ] dx
=∫ (cscx)^2 dx - ∫1/(sinx)^2 d(sinx)
= -cotx+1/sinx

Integral ∫ SiNx √ (1-A ^ 2Sin ^ 2x) DX
RT
In the integral, it is under the root sign of SiNx (1-A square times SiNx Square)

The partial integration method is added with a formula ∫ DX / √ (a ^ 2 + B ^ 2x ^ 2) = (1 / b) ln| BX + √ (a ^ 2 + B ^ 2x ^ 2) | + C

Solving Higher Mathematics integral ∫ e ^ √ (2x + 1) DX
How to calculate? I can't use your method

The answer is (e ^ t) * (t-1), where t = √ (2x + 1)
First, change the element, change √ (2x + 1) to t, simplify and then use the distributed integral method

Calculus d ^ 2x, DX ^ 2, (DX) ^ 2 in Advanced Mathematics
In higher mathematics, what do d ^ 2x, DX ^ 2, (DX) ^ 2 stand for? I've learned them before, but I've forgotten them now
Is there any wrong way of writing or which two have the same meaning?
In fact, I came up with it when I was settling a practical problem. This is the situation
L is a variable and everything else is a function of L
B=l^2;
dx=cosBdl;
da=1/cosBdl;
In this case, I find a relationship, what is the relationship between DX * DA and the differential of L
(I think it's a square relationship)
And then integrate it, ax should be equal to L ^ 2, I don't know if it's correct, if it's correct, what's the correct solution step?
Please help me!
No one answered, so I kept sending it. It's the fourth time today

Super halo! The relationship between the differential of DX * DA and l is of course DX * Da = (DT) ^ 2. But when you say how to integrate it, how can you calculate it like this: ∫ DX * Da = ∫ (DT) ^ 2, it is obviously wrong to deduce XA = T ^ 2. Let me give you a simple example, x = T ^ 3, a = 1 / T, then XA = T ^ 2, but DX = 3T ^ 2DT, Da =

The problem of definite integral of higher numbers: given f (0) = - 1 f (2) = 3 F '(2) = 6, find ∫ XF' '(2x) DX in [0,1]

∫ 1 / (1 + 2x) (1 + x ^ 2) DX how to calculate this?
I know to simplify, but how?

1/(1+2x)(1+x^2)=4/5(1+2x)+(1-2x)/5(1+x^2)
∫1/(1+2x)(1+x^2)dx=∫[4/5(1+2x)+(1-2x)/5(1+x^2)]dx=2/5ln(1+2x)+arctanx-ln(1+x^2)+C

How to calculate ∫ x ^ 3 / (x ^ 2-2x + 3) DX?

The original formula is as follows = 8747; (x ^ 3-3-2x ^ 3-2x ^ 2-2x2x ^ 2-2x ^ 2-2x-2-4x + 6 + 6 (x ^ 2-2x-2-2x-2-2x ^ 2-2x-2-2x ^ 2-2x ^ 2-2x ^ 2-2x ^ 2-2x ^ 2-2x ^ 2-2x2x-2-4x-4x + 6 + 6 + 6 + 6 + 6 + 6 + 6 (x ^ 2-2-2-2x + 3-2x + 3) and (dx-2-2-2x-2x-2x-2x-2x-2x + 2x + 2x + 2x2x + 2x2x2x + 2x + 2x2x + 2x + 2x2x + 2x2x + 2x2x + 2x + 2x2x + 2x + 2x + 2x + 2x + 2x + 2-2x + (x-1) / √ 2) ^ 2 + 1) d ((x-1) / √ 2) = 1 / 2