∫ [e ^ x √ (e ^ x-1)] / (e ^ x + 3) how to calculate DX?

Ans：

Help to calculate ∫ DX / [x + √ (x ^ 2-1)]

If x = sect, then ∫ DX / [x + √ (x ^ 2-1)] = ∫ sect * tantdt / (sect + tant) = ∫ sect * tant (sect tant) DT = ∫ [(sect) ^ 2tant sect (tant) ^ 2] DT = ∫ tantd (tant) - ∫ (sect) ^ 3DT + ∫ sectdt = (1 / 2) (tant) ^ 2 + ln │ sect + tant - ∫ (sect) ^ 3DT calculation ∫ (sect)

Find ∫ (1 / 1-e ^ x) DX

∫(1/1-e^x)dx=∫(1+(e^x)/(1-e^x))dx=x+∫((e^x)/(1-e^x))dx=x-∫(1/1-e^x)d(1-e^x)=x-ln(1-e^x)+C

Horizontal asymptote of function y = (sin (x-1)) / (x ^ 2-x)

The horizontal asymptote is only y = 0
Because LIM (x →∞) y (sin (x-1)) / (x ^ 2-x) = 0
(sin…… Is a bounded function)

Higher number: function y = in (1-x / 1 + x) asymptote
As mentioned above,

Find the asymptote of function y = 1 + 1 / X

X = 0 and y = 1

What is the horizontal asymptote of F (x) = (x ^ 2 + 2) / (x ^ 2-1)?

y=1
When the number of unknowns in the numerator and denominator is equal (twice in this problem), the prime minister coefficient of the numerator is used to compare with the prime minister coefficient of the upper denominator (1 / 1 = 1 in this problem)

Find the asymptote of the following function! Y = (3x) / ((x-3) (x + 2))

Find the asymptote of the following function! Y = (3x) / (x-3) (x + 2)
There are two vertical asymptotes: x = 3; X = - 2
y=3x/(x²-x-6),X→∞lim3x/(x²-x-6)=X→∞lim(3/x)/(1-1/x-6/x²)=0
So there is a horizontal asymptote y = 0 (i.e. X-axis)

Find the asymptote of (x + 2) / (x-1, x + 2) / (x-1)

1)limy=(1+x)/(1+x)=1 (y=1)
2) X ^ 2 + 2x-3 = 0 (x = 1, y = 1, two asymptotes)

The horizontal asymptote of function curve y = x + 1 / X-1 is?
Please tell me the steps of solving the problem in detail. I'm not very good at mathematics! Thank you. By the way, I also want to tell you the difference between horizontal asymptote and vertical asymptote!

Y＝(X-1)/(X+1)=1-2/(x+1)

∫ [e ^ x √ (e ^ x-1)] / (e ^ x + 3) how to calculate DX?