Question21 solve this ASAP

\int \frac{x + 2}{\sqrt{(x - 2) (x - 3)}} dx = \int \frac{x + 2}{\sqrt{(x - 2)} \sqrt{(x - 3)}} dx = \int \frac{x + 2}{\sqrt{x^{2} - 5 x + 6}} dx Writing x + 2 = \frac{1}{2} (2 x - 5) + \frac{9}{2} = \int (\frac{2 x - 5}{2 \sqrt{x^{2} - 5 x + 6}} + \frac{9}{2 \sqrt{x^{2} - 5 x + 6}}) dx = \frac{1}{2} \int \frac{2 x - 5}{\sqrt{x^{2} - 5 x + 6}} dx + \frac{9}{2} \int \frac{1}{\sqrt{x^{2} - 5 x + 6}} dx Consider \int \frac{2 x - 5}{\sqrt{x^{2} - 5 x + 6}} dx Put u = x^{2} - 5 x + 6 \Rightarrow dx = \frac{1}{2 x - 5} du = \int \frac{1}{\sqrt{u}} du = 2 \sqrt{u} = 2 \sqrt{x^{2} - 5 x + 6} + c_{1} Now consider \int \frac{1}{\sqrt{x^{2} - 5 x + 6}} dx = \int \frac{1}{\sqrt{{(x - \frac{5}{2})}^{2} - \frac{1}{4}}} dx (completing the square) Put u = 2 x - 5 \Rightarrow dx = \frac{du}{2} = \int \frac{1}{\sqrt{u^{2} - 1}} du = \ln (\sqrt{u^{2} - 1} + u) + c_{2} = \ln (\sqrt{{(2 x - 5)}^{2} - 1} + 2 x - 5) + c_{2} So, \frac{1}{2} \int \frac{2 x - 5}{\sqrt{x^{2} - 5 x + 6}} dx + \frac{9}{2} \int \frac{1}{\sqrt{x^{2} - 5 x + 6}} dx = \frac{9 \ln (\sqrt{{(2 x - 5)}^{2} - 1} + 2 x - 5)}{2} + \sqrt{x^{2} - 5 x + 6} + C (where C = c_{1} + c_{2})

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