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

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

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Answer 1 · 2018-06-08T16:37:18

See process below

Explanation:

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

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

$x = \frac{- 5 \pm \sqrt{25 + 144}}{12} = \frac{- 5 \pm 13}{6}$ $x=(-5+-sqrt(25+144))/12=(-5+-13)/6$ this give two solutions

$x_{1} = - 3$ $x_1=-3$ and $x_{2} = \frac{4}{3}$ $x_2=4/3$

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

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

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

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

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

1 Answer

Explanation:

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How do you use partial fraction decomposition to decompose the fraction to integrate 266x2+5x−626/(6x^2+5x-6)?

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