Question

please solve this question of mathematics ...class 12... :)​

Answer 1

Answer:

$6\sqrt{x^2-9x+20} +34ln|2x-9 + 2\sqrt{x^2-9x+20}| +c$

Step-by-step explanation:

$\int\limit {\frac{6x+7}{\sqrt{(x-5)(x-4)}} } \, dx$

= $\int\limit {\frac{6x+7}{\sqrt{x^2-9x+20}} } \, dx$    ..............................(i)

y = x²-9x+20

dy = (2x - 9)dx

(i)...................$3\int{\frac{2x-9}{\sqrt{x^2-9x+20}} dx + 34\int{\frac{1}{\sqrt{x^2-9x+20}} dx$

$= 3\int{\frac{dy}{\sqrt{y}}} +34\int{\frac{1}{\sqrt{\frac{1}{4}((2x-9)^2-1)}} dx$

$= 3\times2\sqrt{y} +34\int{\frac{2}{\sqrt{(2x-9)^2-1}} dx$          2x-9 = u  ► 2dx= du

$= 6\sqrt{x^2-9x+20} +34\int{\frac{du}{\sqrt{(u)^2-1}}$

$= 6\sqrt{x^2-9x+20} +34ln|u + \sqrt{u^2-1}|$ +c

$= 6\sqrt{x^2-9x+20} +34ln|2x-9 + \sqrt{(2x-9)^2-1}|$ +c

$= 6\sqrt{x^2-9x+20} +34ln|2x-9 + 2\sqrt{x^2-9x+20}|$ +c