What is #f(x) = int (x+1)/((x+5)(x-4) ) dx# if #f(2)=1 #?

Question

What is #f(x) = int (x+1)/((x+5)(x-4)  )   dx# if #f(2)=1 #?

Emilia Aguilar · Accepted Answer

#f(x) = 1+ (4 ln abs(x+5)+5 ln abs(x-4))/9 - (4 ln 7+5 ln 2)/9#

Decompose the rational function in partial fractions:

#(x+1)/((x+5)(x-4)) = A/(x+5)+B/(x-4)#

#(x+1)/((x+5)(x-4)) = (A(x-4)+B(x+5))/((x+5)(x-4))#

#x+1 = (A+B)x -4A+5B#

#{(A+B=1),(-4A+5B =1):}#

#{(A=4/9),(B=5/9):}#

Then:

#int (x+1)/((x+5)(x-4))dx = 4/9 intdx/(x+5)+5/9 int dx/(x-4)#

#int (x+1)/((x+5)(x-4))dx = 4/9 ln abs(x+5)+5/9 ln abs(x-4)+C#

Now, if #f(2) = 1# then:

#f(x) = 1+int_2^x (t+1)/((t+5)(t-4))dt#

#f(x) = 1+ (4 ln abs(x+5)+5 ln abs(x-4))/9 - (4 ln 7+5 ln 2)/9#

Aurora Alvarado · Answer

undefined To find $ f(x) $, integrate the given function $ \frac{x+1}{(x+5)(x-4)} $ with respect to $ x $. Then use the given condition $ f(2) = 1 $ to solve for the constant of integration.

First, decompose the rational function into partial fractions:

$$ \frac{x+1}{(x+5)(x-4)} = \frac{A}{x+5} + \frac{B}{x-4} $$

Solve for $ A $ and $ B $:

$$ x+1 = A(x-4) + B(x+5) $$
$$ x+1 = (A+B)x + (-4A+5B) $$

Equating coefficients:

$$ A+B = 1 $$
$$ -4A+5B = 1 $$

Solve this system of equations to find $ A $ and $ B $:

$$ A = \frac{-3}{9} $$
$$ B = \frac{4}{9} $$

So, the integral becomes:

$$ \int \left( \frac{-3}{x+5} + \frac{4}{x-4} \right) dx $$

$$ = -3 \ln|x+5| + 4 \ln|x-4| + C $$

Given that $ f(2) = 1 $, substitute $ x = 2 $ and solve for $ C $:

$$ -3 \ln|2+5| + 4 \ln|2-4| + C = 1 $$

$$ -3 \ln(7) + 4 \ln(-2) + C = 1 $$

$$ C = 1 + 3 \ln(7) - 4 \ln(2) $$

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

$$ f(x) = -3 \ln|x+5| + 4 \ln|x-4| + 1 + 3 \ln(7) - 4 \ln(2) $$