Mathematics

$$\displaystyle \int { \dfrac { \left( x+2 \right) dx }{ \sqrt { \left( x-2 \right) \left( x-3 \right)  }  }  }$$ is equal to:


ANSWER

$$\sqrt{x^2-5x+6}+\frac{9}{2}\log\left(x-\frac{5}{2}\right)+\sqrt{x^2-5x+6}$$


SOLUTION
$$\int\frac{x+2.dx}{\sqrt{(x-2)(x-3)}}$$

$$=\int\frac{x+2}{\sqrt{x^2-5x+6}}.dx$$

$$\because x+2=p.(\frac{d}{dx}.x^2-5x+6)+q$$

$$=(2x-5)p+q$$

$$=2px-(5p-q)$$

$$=2p=1 , 5p-q=-2.$$

$$=p=\frac{1}{2},  \frac{5}{2}-q=-2$$

$$=q=\frac{9}{2}$$

$$\int\frac{\frac{1}{2}(2x-5)}{\sqrt{x^2-5x+6}}dx+\int\frac{\frac{9}{2}}{\sqrt{x^2-5x+6}}.dx$$

$$=\frac{1}{2}\int\frac{2x-5}{\sqrt{x^2-5x+6}}.dx+\frac{9}{2}\int\frac{1}{x^2-2.\frac{5}{2}x+\frac{25}{4}-\frac{24}{4+6}}.dx$$

$$=\frac{1}{2}\int\frac{2x-5}{\sqrt{x^2-5x+6}}.dx+\frac{9}{2}\int\frac{1}{\sqrt{(x-\frac{5}{2})^2-(\frac{1}{2}})^2}$$

Let
$$x^2-5x+6=t$$

$$=(2x-5)dx=dt$$

$$=\frac{1}{2}\int \frac{dt}{\sqrt{t}}+\frac{9}{2}\int\frac{1}{\sqrt{{(x-\frac{5}{2})^2}-(\frac{1}{2})^2}}$$

$$=\frac{1}{2}.\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\frac{9}{2}.log[(x-\frac{5}{2})+\sqrt{x^2-5x+6}]$$

$$=\sqrt{x^2-5x+6}+\frac{9}{2}.log(x-\frac{5}{2})+\sqrt{x^2-5x+6}$$

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