What is f(x) = int (x^2+x)/((x+1)(x-1) ) dx if f(3)=2 ?

1 Answer
Apr 20, 2018

f(x) = x+ ln|x-1|-1-ln(2)

Explanation:

Given: f(x) = int (x^2+x)/((x+1)(x-1) ) dx, f(3) = 2

Factor the numerator of the integrand:

f(x) = int (x(x+1))/((x+1)(x-1) ) dx, f(3) = 2

Cancel (x+1)/(x+1) to 1:

f(x) = int x/(x-1) dx, f(3) = 2

Add 0 to the numerator in the form of -1+1:

f(x) = int (x-1+1)/(x-1) dx, f(3) = 2

Separate into two fractions:

f(x) = int (x-1)/(x-1)+1/(x-1) dx, f(3) = 2

The first fraction becomes 1:

f(x) = int 1+1/(x-1) dx, f(3) = 2

The integrals of the two terms are well know:

f(x) = x+ln|x-1|+C, f(3) = 2

Evaluate at x = 3

2 = 3+ln|3-1|+C, f(3) = 2

Solve for C:

C = -1-ln(2)

This is function that satisfies the boundary condition, f(3) = 2:

f(x) = x+ ln|x-1|-1-ln(2)