$\int \frac{2 x - 3}{x^{2} + 3 x - 18} d x$ is equal to :

ln | x^{2} + 3 x - 18 | - \frac{2}{3} ln ∣ ∣ ∣ \frac{x - 3}{x + 6} ∣ ∣ ∣ + C

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ln | x^{2} + 3 x - 18 | + \frac{2}{3} ln ∣ ∣ ∣ \frac{x - 3}{x + 6} ∣ ∣ ∣ + C

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\frac{1}{2} ln | x^{2} + 3 x - 18 | - \frac{2}{3} ln ∣ ∣ ∣ \frac{x - 3}{x + 6} ∣ ∣ ∣ + C

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\frac{3}{2} ln | x^{2} + 3 x - 18 | + \frac{2}{3} ln ∣ ∣ ∣ \frac{x - 3}{x + 6} ∣ ∣ ∣ + C

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Solution

The correct option is A $ln | x^{2} + 3 x - 18 | - \frac{2}{3} ln ∣ ∣ ∣ \frac{x - 3}{x + 6} ∣ ∣ ∣ + C$
Let $2 x - 3 = λ \frac{d}{d x} (x^{2} + 3 x - 18) + μ$
Then $2 x - 3 = λ (2 x + 3) + μ$
Comparing the coefficients of like power of x, we get.
$2 λ = 2$ and $3 λ + μ = - 3$
$\Rightarrow λ = 1$ and $μ = - 6$
$\Rightarrow I = \int \frac{2 x - 3}{x^{2} + 3 x - 18} d x = \int \frac{2 x + 3 - 6}{x^{2} + 3 x - 18} d x = = \int \frac{2 x + 3}{x^{2} + 3 x - 18} d x - 6 \int \frac{1}{x^{2} + 3 x - 18} d x = ln | x^{2} + 3 x - 18 | - 6 \int \frac{1}{x^{2} + 3 x + \frac{9}{4} - \frac{9}{4} - 18} d x = ln | x^{2} + 3 x - 18 | - 6 \int \frac{1}{{(x + \frac{3}{2})}^{2} - {(\frac{9}{2})}^{2}} d x = ln | x^{2} + 3 x - 18 | - 6 \cdot \frac{1}{2 (\frac{9}{2})} ln ∣ ∣ ∣ ∣ ∣ \frac{x + \frac{3}{2} - \frac{9}{2}}{x + \frac{3}{2} + \frac{9}{2}} ∣ ∣ ∣ ∣ ∣ + C = ln | x^{2} + 3 x - 18 | - \frac{2}{3} ln ∣ ∣ ∣ \frac{x - 3}{x + 6} ∣ ∣ ∣ + C$

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