Question

$\int \frac{(x+2)dx}{\sqrt{(x-2)(x-3)}}$ is equal to:

• A. $\sqrt{x^2-5x+6}+\frac{9}{2}\log (x-\frac{5}{2})+\sqrt{x^2-5x+6}$
• B. $\sqrt{x^2-5x+6}-\frac{9}{2}\log (x-\frac{5}{2})+\sqrt{x^2-5x+6}$
• C. $\sqrt{x^2-5x+6}-\frac{9}{2}\log (x-\frac{5}{2})-\sqrt{x^2-5x+6}$
• D. $\sqrt{x^2-5x+6}+\frac{9}{2}\log (x-\frac{5}{2})-\sqrt{x^2-5x+6}$

Answer 1

The correct option is A $\sqrt{x^2-5x+6}+\frac{9}{2}\log (x-\frac{5}{2})+\sqrt{x^2-5x+6}$

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

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

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

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

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

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

$=p=\frac{1}{2},\frac{5}{2}-q=-2$

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

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

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

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

Let

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

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

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

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

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