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

1 Answer
May 1, 2017

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