Two reacting substances which obey the law of mass action starting with 80 kg and 60 kg after chemical reaction leads to the...

We'll integrate to get a logarithmic formula for x and to
determine the amount of reaction product.

Int
dx/(80-x)(60-x) = k Int dt

We'll solve the integral using
the partial fractions method.

We'll write the integrand as
an algebraic sum of irreducible ratios.

1/(80-x)(60-x) =
A/(80-x) + B/(60-x)

We'll calculate  LCD of  the ratios
from the right side:

A/(80-x) + B/(60-x) = [A(60-x) +
B(80-x)]/(80-x)(60-x)

We'll remove the
brackets:

1 = 60A - Ax + 80B -
Bx

We'll combine like terms:

1
= x(-A-B) + 60A + 80B

The coefficients of x from both sides
have to be equal:

-A - B = 0

A
= -B

and free terms:

60A + 80B
= 1

-60B + 80B = 1

20B =
1

B = 1/20

A =
-1/20

1/(80-x)(60-x) = -1/20(80-x) +
1/20(60-x)

Int dx/(80-x)(60-x) = Int-dx/20(80-x) + Int
dx/20(60-x)

We'll substitute 80 - x =
t

We'll differentiate:

-dx =
dt

Int-dx/20(80-x) = Int
dt/t

(Int dt/t)/20 = (ln |t|)/20 +
C

Int-dx/20(80-x) = (ln |80 - x|)/20 +
C

Int dx/20(60-x) = (-ln |60 - x|)/20 +
C

Int dx/(80-x)(60-x) = (ln |80 - x| - ln |60 - x|)/20 +
C

We'll use the quotient property of
logarithms:

Int dx/(80-x)(60-x) = (1/20)*ln
|(80-x)/(60-x)| + C