What is $f (x) = \int \frac{x - 2}{(x + 2) (x - 3)} d x$ $f(x) = int (x-2)/((x+2)(x-3) ) dx$ if $f (2) = - 1$ $f(2)=-1$ ?

Question

1470 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-03T20:43:08

$f (x) = \frac{4}{5} ln | x + 2 | - \frac{1}{5} ln | x - 3 | - 1 - \frac{4}{5} ln (4)$ $f(x)=4/5lnAbs(x+2)-1/5lnAbs(x-3) -1-4/5ln(4)$

Begin by splitting the fraction into partial fractions:

$\frac{x - 2}{(x + 2) (x - 3)} = \frac{A}{x + 2} + \frac{B}{x - 3}$ $(x-2)/((x+2)(x-3)) = A/(x+2) + B/(x-3)$

Now multiply by $(x + 2) (x + 3)$ $(x+2)(x+3)$ to find $A$ $A$ and $B$ $B$ .

$x - 2 = (x - 3) A + (x + 2) B$ $x-2 = (x-3)A + (x+2)B$

Set $x = 3$ $x=3$ to find $B$ $B$ .

$(3) - 2 = ((3) - 3) A + ((3) + 2) B = 5 B \to 5 B = 1$ $(3)-2 = ((3)-3)A+ ((3)+2)B = 5B -> 5B =1$
$therefore B =1/5$

Set $x = -2$ to find $A$ so we will have:
$-4 =-5A therefore A = 4/5$

So we can now rewrite the integral in partial fractions and integrate:

$int (x-2)/((x+2)(x-3))dx = int 4/(5(x+2)) + 1/(5(x-3))dx$

$therefore f(x)=4/5lnAbs(x+2)-1/5lnAbs(x-3) + C$

Now you can use the condition specified: $f(2) = -1$

$-> -1 = 4/5lnAbs(4) -1/5lnAbs(-1) + C$

$lnAbs(-1) = ln(1) = 0$ which leaves us with:

$C = -1-4/5ln(4)$ which means our final answer is:

$f(x)=4/5lnAbs(x+2)-1/5lnAbs(x-3) -1-4/5ln(4)$

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