how to integrate 1/y(1-y)? show it step by step?

Question

the answer is is -ln( (y-1)/y)

Answer 1

You have to use partial fractions method ..RHS= A/Y + B/(1-Y)

The answer Could be A Ln Y + B Ln (1-Y) + C

Where A,B can be found and C is a Constant too..

Answer 2

So you want to solve

Integral (1 / [y(1 - y)])dy

Your first step is to use partial fractions to make the integral solveable.

We can rewrite the fraction 1/[y(1 - y)] as follows:

1/[y(1 - y)] = A/y + B/(1 - y)

Since we have linear factors, we can decompose the fraction in this manner. We have to solve for A and B. We can either solve for A and B before or after solving for the integral; I choose to solve for them after.

Integral (1 / [y(1 - y)])dy = Integral (A/y + B/(1 - y))dy

We can split this up into two separate integrals.

Integral (A/y)dy + Integral (B/(1-y))dy

We can also pull the constants out of the integrals.

A * Integral (1/y) dy + B * Integral (1/(1 - y))dy

These are easy integrals to solve for. Note that the second integral is *like* ln|1 - y| EXCEPT for the fact that taking the derivative would result in using the chain rule and multiplying (-1). To offset this, we have to include (-1) in our integral. Thus, we get:

A*ln |y| + B*(-ln|1 - y|) + C, OR

A*ln|y| - B*ln|1 - y| + C

Now, let's solve for A and B. Recall that

1/[y(1 - y)] = A/y + B/(1 - y)

If we multiply both sides by y(1 - y), we can remove ALL fractions, and obtain

1 = A(1 - y) + B(y). Expanding the right hand side,
1 = A - Ay + By. Grouping together the y terms, we get
1 = A + By - Ay
1 = A + (B - A)y.

Now, just to make something obvious, I'm going to add 0y to the right hand side. After all, 0y = 0, and adding 0 does nothing to an equation

1 + 0y = A + (B - A)y

By the method of partial fractions, this means we can equate the left hand side and right hand side component-wise. Notice on the left hand side, we have a constant term, and a coefficient of y. We have this on the right hand side too, and what we do is EQUATE them.

1 = A
0 = B - A

We solve this system of equations and get A = 1 and B = 1, so we're finished solving for A and B.

Recall that our solution was

A*ln|y| - B*ln|1 - y| + C

But now that we solved for A and B, and got A = 1 and B = 1, we substitute appropriately.

(1) ln|y| - (1)ln|1 - y| + C
ln|y| - ln|1 - y| + C

Remember that, by the log property
log[base b](a/c) = log[base b](a) - log[base b](c), we can combine those two logs into a single log. Our final answer should be:

ln |(y)/(1 - y)| + C

Answer 3

To solve the problem use the method of partial fraction

1/y(1-y) = A/y + B/ (1 –y )
A(1-y)+By =1
A-Ay + By =1
A + (-A + B)y = 1
Compare the coefficients of y and constant term
A = 1
-A + B = 0
A = B = 1


1/y(1-y) = 1/y + 1/1-y
Integral 1/y(1-y) = integral (1/y) dy + integral (1/1-y)
= ln y – ln (1 - y) + C

Answer 4

use integration by partial fraction.,

A / y + B/( 1-y)

combining

A - Ay + By / y (1-y)

y: 0 = -A + B
1: 1 = A

this implies that B = 1

so integrate 1/y + 1/ (1-y)

this is equal to ln | y | - ln | 1 - y | + C

this is also equal to ln ( y / (1- y )) + C., done

Answer 5

First decompose the fraction.

1/[y(1 - y) = 1/y + 1/(1-y)

Now integrate.

∫[1/y + 1/(1-y)]dy = ln |y| - ln |1-y| + C = ln |y/(1-y)| +C

As Puggy rightly notes, the value in the natural log is an absolute value. This is because you can't take a log of a negative number.

Answer 6

I think this answer satisfies u
first step; find the partial fractions
1/y(1-y)=a/y+b/1-y
1=a(1-y)+by
1=a-ay+by
compare constants
a=1
compare y coeffecients
0=-a+b
a=b
we have a=1
so, b=1
1/y(1-y)=1/y+1/1-y
apply integration
integral 1/y(1-y)=integral 1/y+ integral 1/1-y
= ln|y| +[-ln|1/1-y| ]
= lny-ln1/1-y
= ln(y/1-y)
= ln(1-y/y)-1
= -ln(1-y/y)
hence 1/y(1-y)= -ln(1-y/y)

Answer 7

You use partial fractions, c'mon I learned this in a second without ever taking calculus 1/x(x - 4) = A/(x - 4) + B(x) 1 = x(A + B) - 4B A = 1/4 and B = -1/4 = 1/4 *ln(x - 4) - 1/4*ln(x) + k

Answer 8

1/y(1 - y) = A/y + B/(1 - y)
A = 1
B = 1
∫[1/y(1-y)]dy =
∫dy/y + ∫dy/(1 - y) =
ln(y) - ln(1 - y) =
ln(y/(1 - y)) + C
ln(y/(1 - y)) = -ln((1 - y)/y)

Answer 9

resolve into patrtial fractions and integrate
A/y+B/1-y=1/y(1-y)
A(1-y)+By=1
A-Ay+By=1
A=1
B=-1

so 1/y-1/1-y
lny-ln(1-y)+C