It is given f(x)=(19x-2)/((5-x)(1+6x)) can be expressed A/(5-x)+B/(1+6x) where A and B are integers. i) Find A and B ii) Show the integral of this from 0 to 4 = Kln5

Firstly, we are given that f(x) can be expressed in the above form, so we write this out:

(19x-2)/((5-x)(1+6x)=A/(5-x)+B/(1+6x)

We then multiply by the denominator of f(x):

19x-2=A(1+6x)+B(5-x)

Now we can choose values of x such that each of the brackets equal 0 to find A and B.

x=5  95-2=31A  A=3

x=-1/6  -31/6=(31/6)B  B=-1

So we can write f(x)=3/(5-x)-1/(1+6x)

Now part ii) we can replace the f(x) in the integral with this:

integral(3/(5-x)-1/(1+6x))

We can separate this into

integral(3/(5-x))-integral(1/(1+6x))

Now we want to make the numerator the derivative of the denominator from the form of the answer we're looking for:

-3integral(-1/(5-x))-(1/6)integral(6/(1+6x))

Which equals

-3ln(5-x)-(1/6)ln(1+6x) 

We can sub the limits straight into this:

-3ln1-(1/6)ln25-(-3ln5-(1/6)ln1))

We know ln1=0 so we have

-(1/6)ln25+3ln5

We can rewrite ln25 as 2ln5 to give

(-1/3)ln5+3ln5= (8/3)ln5

i.e. K=8/3