What is f(x) = int (x+1)/((x+5)(x-1) ) dx if f(1)=6 ?
1 Answer
The general form is:
f(x) = 2/3 ln|x+5| + 1/3ln|x-1| + lnC
There is no such
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) = 1/3 { 2ln6 + ln0 }+ lnC
And
The general form is:
f(x) = 2/3 ln|x+5| + 1/3ln|x-1| + lnC
There is no such