How do you integrate $\int \frac{13 x}{6 x^{2} + 5 x - 6}$ $int (13x) / (6x^2 + 5x - 6)$ using partial fractions?

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How do you integrate $\int \frac{13 x}{6 x^{2} + 5 x - 6}$ $int (13x) / (6x^2 + 5x - 6)$ using partial fractions?

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Answer 1 · 2016-06-14T23:39:52

$\frac{2}{3} ln (| 3 x - 2 |) + \frac{3}{2} ln (| 2 x + 3 |) + C$ $2/3ln(abs(3x-2))+3/2ln(abs(2x+3))+C$

Explanation:

Note that

$6 x^{2} + 5 x - 6 = 6 x^{2} + 9 x - 4 x - 6$ $6x^2+5x-6=6x^2+9x-4x-6$

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

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

The partial fraction decomposition can then be set up as:

$\frac{13 x}{(3 x - 2) (2 x + 3)} = \frac{A}{3 x - 2} + \frac{B}{2 x + 3}$ $(13x)/((3x-2)(2x+3))=A/(3x-2)+B/(2x+3)$

Multiplying both sides by $(3 x - 2) (2 x + 3)$ $(3x-2)(2x+3)$ , we see that

$13 x = A (2 x + 3) + B (3 x - 2)$ $13x=A(2x+3)+B(3x-2)$

Letting $x = - \frac{3}{2}$ $x=-3/2$ :

$13 (- \frac{3}{2}) = A (2 (- \frac{3}{2}) + 3) + B (3 (- \frac{3}{2}) - 2)$ $13(-3/2)=A(2(-3/2)+3)+B(3(-3/2)-2)$

$- \frac{39}{2} = A (- 3 + 3) + B (- \frac{9}{2} - 2)$ $-39/2=A(-3+3)+B(-9/2-2)$

$- \frac{39}{2} = B (- \frac{13}{2})$ $-39/2=B(-13/2)$

$3 = B$ $3=B$

Letting $x = \frac{2}{3}$ $x=2/3$ :

$13 (\frac{2}{3}) = A (2 (\frac{2}{3}) + 3) + B (3 (\frac{2}{3}) - 2)$ $13(2/3)=A(2(2/3)+3)+B(3(2/3)-2)$

$\frac{26}{3} = A (\frac{4}{3} + 3) + B (2 - 2)$ $26/3=A(4/3+3)+B(2-2)$

$\frac{26}{3} = A (\frac{13}{3})$ $26/3=A(13/3)$

$2 = A$ $2=A$

Thus,

$\frac{13 x}{(3 x - 2) (2 x + 3)} = \frac{2}{3 x - 2} + \frac{3}{2 x + 3}$ $(13x)/((3x-2)(2x+3))=2/(3x-2)+3/(2x+3)$

So,

$\int \frac{13 x}{6 x^{2} + 5 x - 6} d x = 2 \int \frac{1}{3 x - 2} d x + 3 \int \frac{1}{2 x + 3} d x$ $int(13x)/(6x^2+5x-6)dx=2int1/(3x-2)dx+3int1/(2x+3)dx$

$= \frac{2}{3} \int \frac{3}{3 x - 2} d x + \frac{3}{2} \int \frac{2}{2 x + 3} d x$ $=2/3int3/(3x-2)dx+3/2int2/(2x+3)dx$

$= \frac{2}{3} ln (| 3 x - 2 |) + \frac{3}{2} ln (| 2 x + 3 |) + C$ $=2/3ln(abs(3x-2))+3/2ln(abs(2x+3))+C$

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How do you integrate $\int \frac{13 x}{6 x^{2} + 5 x - 6}$ $int (13x) / (6x^2 + 5x - 6)$ using partial fractions?

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How do you integrate int (13x) / (6x^2 + 5x - 6)∫13x6x2+5x−6int (13x) / (6x^2 + 5x - 6) using partial fractions?

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How do you integrate $\int \frac{13 x}{6 x^{2} + 5 x - 6}$ $int (13x) / (6x^2 + 5x - 6)$ using partial fractions?