Question

Im struggling with the problem Int[dy/(1+y)^2] = Int[dx/(1+x)^2

Answer 1

<Z>I<z>nt[(dy)/((1+y)^(2))]=<Z>I<z>nt[(dx)/((1+x)^(2))]

Divide each term in the equation by <Z>I<z>nt.
(<Z>I<z>nt[(dy)/((1+y)^(2))])/(<Z>I<z>nt)=(<Z>I<z>nt[(dx)/((1+x)^(2))])/(<Z>I<z>nt)

Simplify the left-hand side of the equation by canceling the common factors.
(dy)/((1+y)^(2))=(<Z>I<z>nt[(dx)/((1+x)^(2))])/(<Z>I<z>nt)

Simplify the right-hand side of the equation by simplifying each term.
(dy)/((1+y)^(2))=(dx)/((1+x)^(2))

Since there is one rational expression on each side of the equation, this can be solved as a ratio. For example, (A)/(B)=(C)/(D) is equivalent to A*D=B*C.
dy*(1+x)^(2)=dx*(1+y)^(2)

Multiply dy by (1+x)^(2) to get dy(1+x)^(2).
dy(1+x)^(2)=dx*(1+y)^(2)

Multiply dx by (1+y)^(2) to get dx(1+y)^(2).
dy(1+x)^(2)=dx(1+y)^(2)

Since dx(1+y)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting dx(1+y)^(2) from both sides.
dy(1+x)^(2)-dx(1+y)^(2)=0

Squaring an expression is the same as multiplying the expression by itself 2 times.
(dy*(1+x)(1+x))-dx(1+y)^(2)=0

Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.
(dy(1*1+1*x+x*1+x*x))-dx(1+y)^(2)=0

Simplify the FOIL expression by multiplying and combining all like terms.
(dy(x^(2)+2x+1))-dx(1+y)^(2)=0

Multiply dy by each term inside the parentheses.
(dx^(2)y+2dxy+dy)-dx(1+y)^(2)=0

Squaring an expression is the same as multiplying the expression by itself 2 times.
(dx^(2)y+2dxy+dy)+(-dx*(1+y)(1+y))=0

Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.
(dx^(2)y+2dxy+dy)+(-dx(1*1+1*y+y*1+y*y))=0

Simplify the FOIL expression by multiplying and combining all like terms.
(dx^(2)y+2dxy+dy)+(-dx(y^(2)+2y+1))=0

Multiply -dx by each term inside the parentheses.
(dx^(2)y+2dxy+dy)+(-dxy^(2)-2dxy-dx)=0

Remove the parentheses that are not needed from the expression.
dx^(2)y+2dxy+dy-dxy^(2)-2dxy-dx=0

Since 2dxy and -2dxy are like terms, add -2dxy to 2dxy to get 0.
dx^(2)y+dy-dxy^(2)-dx=0

Reorder the polynomial dx^(2)y+dy-dxy^(2)-dx alphabetically from left to right, starting with the highest order term.
dx^(2)y-dxy^(2)+dy-dx=0

Use the quadratic formula to find the solutions. In this case, the values are a=dx^(2), b=-1dx, and c=d-dx.
y=(-b\~(b^(2)-4ac))/(2a) where ay^(2)+by+c=0

Use the standard form of the equation to find a, b, and c for this quadratic.
a=dx^(2), b=-1dx, and c=d-dx

Substitute in the values of a=dx^(2), b=-1dx, and c=d-dx.
y=(-(-1dx)\~((-1dx)^(2)-4(dx^(2))(d-dx)))/(2(dx^(2)))

Remove the -1 from -1dx.
y=(-(-dx)\~((-1dx)^(2)-4(dx^(2))(d-dx)))/(2(dx^(2)))

Multiply -1 by each term inside the parentheses.
y=(dx\~((-1dx)^(2)-4(dx^(2))(d-dx)))/(2(dx^(2)))

Simplify the section inside the radical.
y=(dx\dx~(-1(3-4x)))/(2(dx^(2)))

Simplify the denominator of the quadratic formula.
y=(dx\dx~(-1(3-4x)))/(2dx^(2))

First, solve the + portion of \.
y=(dx+dx~(-1(3-4x)))/(2dx^(2))

Simplify the expression to solve for the + portion of the \.
y=(1+~(-1(3-4x)))/(2x)

Next, solve the - portion of \.
y=(dx-dx~(-1(3-4x)))/(2dx^(2))

Simplify the expression to solve for the - portion of the \.
y=(1-~(-1(3-4x)))/(2x)

The final answer is the combination of both solutions.
y=(1+~(-1(3-4x)))/(2x),(1-~(-1(3-4x)))/(2x)

Answer 2

I believe I'm supposed to be taking the integral of both sides.
The answer the book gives is y=(1+x)/[1+C(1+x)]-1

Answer 3

mhhhmmm, really?
Did it ask you to find what or do what?

Answer 4

Its just looking for me to solve for y.

Thanks

Answer 5

Wow, that's huge. @some_someone

Answer 6

lolol ikr?

Answer 7

lolz :)

Answer 8

i just want them to fully understand :D

Answer 9

I still can't believe it's that long.. D:

Answer 10

Sometimes I type my work out in a text document before actually pasting it into the thread. It makes it much faster to type when I don't have the LaTeX generated in real time as I type.

So I can certainly see how it might seem like he's taking the info from somewhere else. But he's likely just typing it out, taking his time.

Answer 11

@JENFALYONS were you able to figure this problem out?

Answer 12

lets concentrate on the problem ,
so, you need, \(\int \dfrac{1}{(1+y)^2} dy\)
put u= 1+y
then du=..... ?

Answer 13

please do/continue such conversations in private messages.
i suggest you delete all unnecessary comments.

Answer 14

now waiting for @JENFALYONS response/doubts...