What is LIM (x tends to + ∞) ∫ (0 → x) (arctant) & # 178; DT / √ (1 + X & # 178;) equal to?

What is LIM (x tends to + ∞) ∫ (0 → x) (arctant) & # 178; DT / √ (1 + X & # 178;) equal to?

lim（x→+∞）∫（0→x）（arctant）²dt/√（1+x²）
=lim（x→+∞）∫（0→x）（arctant）²dt/x (∞/∞)
=lim（x→+∞）（arctanx）²
=π^2/4

Solving Lim [∫ superscript x subscript 0 (arctant) &# 178; DT] / √ X & # 178; + 1, X tends to + ∞

When x tends to + ∞ by using the law of Robita, the numerator: ∫ superscript x, subscript 0 (arctant) &# 178; DT denominator: √ x &# 178; + 1, the numerator and denominator tend to be positive infinity. Therefore, the limit value is equal to the ratio of the upper and lower to the X derivative when x tends to + ∞. The numerator's derivative to x = (arctant) &# 178; the denominator's derivative to x = x / √ (x &# 178

Why is LIM (x tends to ∞) 2x / X-1 equal to E & # 178;?

2x/x-1=2(x-1)/x-1+2/x-1
When x approaches infinity, 2 / X-1 = 0
So there is the above conclusion

What does LIM (x tends to + ∞) ∫ (0 → x) (2arctantdt) / √ (1 + X & # 178;) equal?

Solution 1: ∫ 2arctantdt = 2xarctanx-2 ∫ TDT / (1 + T & # 178;) (Application of partial integration method)
=2xarctanx-ln(1+x²)
LIM (x - > + ∞) [ln (1 + X & # 178;) / x] = LIM (x - > + ∞) [2x / (1 + X & # 178;)] (∞ / ∞ type limit, applying the law of Robida)
=lim(x->+∞)[(2/x)/(1+1/x²)]
=0
The original formula = LIM (x - > + ∞) [(2xarctanx ln (1 + X & # 178;) / √ (1 + X & # 178;)]
=LIM (x - > + ∞) [(2arctanx ln (1 + X & # 178;) / x) / √ (1 + 1 / X & # 178;)] (numerator and denominator divide by x)
=[2(π/2)-0]/√(1+0)
=π；
Solution 2: the original formula = LIM (x - > + ∞) [2arctanx / (x / √ (1 + X & # 178;)] (∞ / ∞ type limit, applying the law of Robida)
=2[lim(x->+∞)(arctanx)]*{lim(x->+∞)[√(1+1/x²]}
=2(π/2)*√(1+0)
=π.

Find the indefinite integral ∫ ln (T + a) DT, a is a constant

[Note: using the "step-by-step integration method] ∫ ln (x + a) DX = ∫ ln (x + a) d (x + a) = (x + a) ln (x + a) -∫ (x + a) d [ln (x + a)] = (x + A) ln (x + a) -∫ DX = (x + a) ln (x + a) - x + C. ∫ ln (x + a) DX = [(x + a) ln (x + a)] - x + C

A problem about definite integral, such as finding a derivative D ∫ [0, x] e ^ 2T DT / DX
Why can't we use Leibniz formula to find ∫ [0, x] e ^ 2T DT = e ^ 2x-e ^ 0 = e ^ 2x-1 first
Is it wrong to find the derivative of x = 2E ^ 2x?

∫[0,X] e^2t dt=1/2*∫[0,X] e^2t d(2t)=1/2*[e^(2x)-1]
So you got the definite integral wrong, 1 / 2 less

Finding the indefinite integral of T ^ 100 / (t-1) DT

t^100/(t-1)
=(t^100-1+1)/(t-1)
=(t^100-1)/(t-1)+1/(t-1)
And T ^ 100-1 = (t-1) (T ^ 99 + T ^ 98 + T ^ 97 +...) +t+1)
So (T ^ 100-1) / (t-1) = T ^ 99 + T ^ 98 + T ^ 97 + +t+1
For the right integral: T ^ 100 / 100 + T ^ 99 / 99 + +t^2/2+t+C1
The integral of 1 / (t-1) is ln (t-1) + C2
So the integral of the original formula is: T ^ 100 / 100 + T ^ 99 / 99 + +t^2/2+t+C1+ln(t-1)+C2
=t^100/100+t^99/99+…… +t^2/2+t+ln(t-1)+C
= ∑(i,n)t^i/i+ln(t-1)+C

What is the solution of indefinite integral ∫ (T / 1 + T) DT?
Why not?
∫ (T / 1 + T) DT = ∫ DT - ∫ 1 / (1 + T) DT = t - ∫ 1 / (1 + T) d (T + 1) = T-ln T + 1 + C

There is no problem. It should be solved in this way~

What is the indefinite integral of T * e ^ (T ^ 2)
Ask for math, thank you

t*e^(t^2)dt=1/2*e^(t^2)d(t^2)=1/2d[e^(t^2)]
Indefinite integral of T * e ^ (T ^ 2) = 1 / 2 * e ^ (T ^ 2) + C

The Laplace transform of COS (WT) needs only a specific formula

From Euler formula
cos(wt)=(1/2)*[e^iwt+e^(-iwt)]
L(coswt)=(1/2)L[e^iwt+e^(-iwt)]
=(1/2)*[L(e^iwt)+L(e^-iwt)]
And l (e ^ at) = 1 / (S-A)
So the original formula = (1 / 2) [1 / (s-iw) + 1 / (s + IW)]
=s/(s^2+w^2)