How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{11}{3 x^{2} - 7 x - 6}$ $11/(3x^2-7x-6)$ ?

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Answer 1 · 2018-04-04T08:48:20

$\int \frac{11}{3 x^{2} - 7 x - 6} d x = ln ∣ ∣ ∣ \frac{x - 3}{3 x + 2} ∣ ∣ ∣ + C$ $int 11/(3x^2-7x-6)dx = ln abs((x-3)/(3x+2))+C$

Factorize the denominator by solving the equation:

$3 x^{2} - 7 x - 6 = 0$ $3x^2-7x-6 = 0$

$x = (7+-sqrt(49+72))/6 = (7+-sqrt(121))/6 = (7+-11)/6={(-2/3),(3):}$

Then:

$3x^2-7x-6 = 3(x+2/3)(x-3) = (3x+2)(x-3)$

Apply partial fractions decomposition:

$11/(3x^2-7x-6) = A/(3x+2)+B/(x-3)$

$11/(3x^2-7x-6) = (A(x-3)+B(3x+2))/(3x^2-7x-6)$

$11 = (A+3B)x - 3A +2B$

${(A+3B = 0),(-3A+2B=11):}$

${(A=-3),(B=1):}$

So:

$11/(3x^2-7x-6) = -3/(3x+2)+1/(x-3)$

$int 11/(3x^2-7x-6)dx = -int (3dx)/(3x+2)+int dx/(x-3)$

$int 11/(3x^2-7x-6)dx = -int (d(3x+2))/(3x+2)+int (d(x-3))/(x-3)$

$int 11/(3x^2-7x-6)dx = -ln abs(3x+2)+ln abs(x-3)+C$

$int 11/(3x^2-7x-6)dx = ln abs((x-3)/(3x+2))+C$

How do you use partial fraction decomposition to decompose the fraction to integrate 11/(3x^2-7x-6)113x2−7x−611/(3x^2-7x-6)?