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

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Answer 1 · 2018-04-20T17:42:29

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

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)#