What is $f (x) = \int \frac{x + 1}{(x + 5) (x - 4)} d x$ $f(x) = int (x+1)/((x+5)(x-4) ) dx$ if $f (2) = 1$ $f(2)=1$ ?

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Answer 1 · 2018-05-07T13:52:03

$f (x) = 1 + \frac{4 ln | x + 5 | + 5 ln | x - 4 |}{9} - \frac{4 ln 7 + 5 ln 2}{9}$ $f(x) = 1+ (4 ln abs(x+5)+5 ln abs(x-4))/9 - (4 ln 7+5 ln 2)/9$

Decompose the rational function in partial fractions:

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

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

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

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

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

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:

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

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

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