How do you use partial fraction decomposition to decompose the fraction to integrate $26/(6x^2+5x-6)$ ?

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Answer 1 · 2018-06-08T16:37:18

See process below

First of all, we factorize the polynomial $6x^2+5x-6$

$6x^2+5x-6=0$

$x=(-5+-sqrt(25+144))/12=(-5+-13)/6$ this give two solutions

$x_1=-3$ and $x_2=4/3$

Thus we have $6x^2+5x-6=(x+3)(x-4/3)$

$1/(6x^2+5x-6)=A/(x+3)+B/(x-4/3)$ transposing terms and equalizing we have

$A+B=0$
$3B-4/3A=1$ from here we find $A=-3/13$ and $B=3/13$

Then finally we have $1/(6x^2+5x-6)=(-3/13)/(x+3)+(3/13)/(x-4/3)$

How do you use partial fraction decomposition to decompose the fraction to integrate 26/(6x^2+5x-6)26/(6x^2+5x-6)?