Question

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$\int \: \frac{9x + 2}{ {x}^{2} + x - 6 } dx$

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Answer 1

Solution :

$\tt \implies \:( Let ) \: \: I = \int \dfrac{9x + 2}{ {x}^{2} + x - 6 } dx$

$\tt \implies \: I = \int \dfrac{9x + 2}{ (x + 3)(x - 2) } dx$

Form of the Partial fraction.

$\tt \implies \: \dfrac{9x + 2}{ (x + 3)(x - 2) } = \frac{A}{(x - 2)} + \frac{B}{(x + 3)}$

$\tt \implies \: \dfrac{9x + 2}{ (x + 3)(x - 2) } = \frac{A(x + 3) + B(x - 2)}{(x + 3)(x - 2)}$

$\tt \implies \: 9x + 2 = A(x + 3) + B(x - 2) \\ \\ \tt- - - - (1)$

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From ( 1 )

At,

• x = 2.

$\tt \implies \: 20 = 5A. \:$

$\tt \implies \: 4= A \:$

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At,

• x = - 3.

$\tt \implies \: - 25 = - 5B$

$\tt \implies \: 5 = B$

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Now,

$\tt \implies \: I = \int \dfrac{(9x + 2 ) dx}{ {x}^{2} + x - 6 }$

$\tt \implies I \: = \: \: \int \frac{A \: dx}{(x - 2)} + \int\frac{B \: dx}{(x + 3)}$

$\tt \implies I \: = \: \: \int \frac{4 \: dx}{(x - 2)} + \int\frac{5\: dx}{(x + 3)}$

$\tt \implies I \: = \: \: 4 \int \frac{dx}{(x - 2)} + 5 \int\frac{dx}{(x + 3)}$

$\tt \implies I \: = \: \: 4 \: ln |x - 2| + 5 \: ln | x + 3 | + c.$