alright so i got A = 1 and B = 2 so now i have the integral of 1/u + 2/(1-u). Which i end up getting ln |u| - 2ln|1-u| + c. Now replacing the u for e^x i get ln e^x - 2 ln(1-e^X). Which is x-2 ln (1-e^x) + c. Is that the right answer?
Homework Statement
integral of (1+e^x)/(1-e^x) dx
Homework Equations
The Attempt at a Solution
The TA said to make u = e^x
So, du = e^x dx. dx = du/e^x.
Since e^x = u
The integral now is (1+u)/(1-u)u
I am confused as to what to do after distribute the u in the bottom.
After i complete the square i end up with dx/((x+2)^2+1)^2. I substitute u = x+2 , du = dx
so i get the integral du/(u^2+1)^2. Then i got u = tan theta, du = sec^2(theta) d(theta) and i replace that in the integral as sec^2(theta) d(theta)/(tan^2(theta) +1)^ and i get after i substitute sec^2...
oh, i see what you mean. Basically if there is only one type of polynomial squared or cubed in the denominator then i can't use partial fraction but if suppose there is (x+2)(x-1)^2 in the denominator then i could use partial fractions and expand the as a/(x+2) + b/(x-1) + c/(x-1)^2 right? Also...
so when there is only dx sitting on the top i cannot use partial fraction decomposition? How can i tell if its already in expanded form. On the directions of this problem the TA said to use PFD but it clearly does not work.
Homework Statement
\int 1/(x^{2}+4x+5)^{2}
Homework Equations
I am using partial fraction decomposition
The Attempt at a Solution
1/(x^{2}+4x+5)^{2} = Ax+b/(X^2+4x+5) + Cx+D/(x^2+4x+5)^2
1 = (Ax+b)(X^2+4x+5) + Cx+D
When i multiply through to find the values for A and b...
maybe I am not understanding your answer correctly. This is what i have for y so far. y = 2(b/2 +(a/2-b/2)*y/h). I put the two outside so whatever the x value comes out to be it will be multiplied by 2 that way i get the whole length and not just half of it. If h = 0 then y = 0 , thus we are...
Homework Statement
Find the volume of the frustum of a pyramid with square base of side b,
square top of side a, and height h. What happens if a = b ? If a = 0 ?
Homework Equations
None
The Attempt at a Solution
I know that i have to integrate from 0 to H. I make a generic...
62.4 is the density of water the teacher wanted us to use.I decided to keep all my numbers positive which will also make it much easier to integrate.Here is my new integral with expressing r as a function of y. 62.4*pi * integral from 0 to 5 of ((5/4y)^2) * (8-y)
8-y is the distance the water...
Homework Statement
Find the work done in pumping all the water out of a conical reservoir of radius 10ft at the top and altitude 8ft if at the beginning the reservoir is filled to a depth of 5ft and the water is pumped just to the top of the reservoir.
Homework Equations
None
The Attempt at a...
Thanks for correcting me i always say pie for some reason. This problem seemed pretty difficult but drawing it does make a lot of difference..Thanks for your help
i worked the problem a little further. Since the D= 2x i would get (pie (D/2)^2)/2. So it would be (pie * D^2)/8 and diameter is 2x so that would give me pie * x^2/2. Then i get (pie * X^2)/2 and since X^2 + Y^2 = 9 is the circles formula X^2 would be (9-y^2). i pulled out pie/2 outside of the...