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

1 Answer
Steve M
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#

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