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

Question

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

Charles Arocha · Accepted Answer

#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)#

Charles Anctil · Answer

undefined To find $ f(x) = \int \frac{x^2+x}{(x+1)(x-1)} \, dx $ if $ f(3) = 2 $, first, compute the indefinite integral:

$$ \int \frac{x^2+x}{(x+1)(x-1)} \, dx = \int \left(\frac{1}{x-1} - \frac{1}{x+1}\right) \, dx $$

$$ = \ln|x-1| - \ln|x+1| + C $$

Given that $ f(3) = 2 $, substitute $ x = 3 $ and $ f(x) = 2 $ into the equation:

$$ 2 = \ln|3-1| - \ln|3+1| + C $$

$$ 2 = \ln 2 - \ln 4 + C $$

$$ 2 = \ln\frac{1}{2} + C $$

$$ C = 2 - \ln\frac{1}{2} $$

$$ C = 2 + \ln 2 $$

So, the function $ f(x) $ is:

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