Question

What is `f(x) = int (x+1)/((x+5)(x-1) ) dx` if `f(1)=6`?

Answer 1

The general form is:

`f(x) = 2/3 ln|x+5| + 1/3ln|x-1| + lnC`

There is no such `f(x)` satisfying the given condition `f(1)=6`

Explanation:

We can decompose the integrand into partial fractions as follows:

`(x+1)/((x+5)(x-1)) -= A/(x+5) + B/(x-1)`
`" " = (A(x-1) + B(x+5))/((x+5)(x-1))`

`:. x+1 -= A(x-1) + B(x+5)`

Put:

`x=-5 => -5+1=A(-5-1) => A=2/3`
`x= \ \ \ \ 1 => \ \ \ \ \ \1+1=B(1+5) \ \ \ \ \ => B=1/3`

Thus, we can write:

`f(x) = int \ (x+1)/((x+5)(x-1)) \ dx`
`" " = int \ (2/3)/(x+5) + (1/3)/(x-1) \ dx`
`" " = 1/3 \ int \ (2)/(x+5) + 1/(x-1) \ dx`
`" " = 1/3 { 2ln|x+5| + ln|x-1| }+ lnC`
`" " = 2/3 ln|x+5| + 1/3ln|x-1| + lnC`

We are given that `f(1) = 6`; which causes an issue;

`f(1) = 1/3 { 2ln6 + ln0 }+ lnC`

And `ln 0` is undefined:

The general form is:

`f(x) = 2/3 ln|x+5| + 1/3ln|x-1| + lnC`

There is no such `f(x)` satisfying the given condition `f(1)=6`