Question

Let F(x) be the primitive of $\frac{3x+2}{\sqrt{x-9}}$ w.r.t. $x$. If $F\left(10\right)=60$ then the value of $F\left(13\right)$, is

• A. $66$
• B. $132$
• C. $248$
• D. $264$

Answer 1

The correct option is B $132$

$I\text{=}\sqrt{\frac{3x+2}{x-9}}dx$

$\sqrt{x-9}=t$

$x-9=t^2\Rightarrow \text{x =}{\text{t}}^2+9$

$dx=2tdt$

$I=\sqrt{\frac{3\left(t^2+9\right)+2}{t}}\times 2tdt$

$\text{= 2}\sqrt{\frac{3t^2+29}{t}}\times dt$

$\text{= 2}\left(3\frac{t^3}{3}+29t\right)+c$

$f\left(x\right)\text{=}2\left({\left(x-9\right)}^{3/2}+29{\left(x-9\right)}^{1/2}\right)+c$

$f\left(10\right)=60$

$f10=2((1{)}^{\frac{3}{2}}+29{\left(1\right)}^{\frac{1}{2}})+c=60$

$\text{60 + c = 60}$

$\text{c = 0}$

$f\left(x\right)=2\left({\left(x-9\right)}^{\frac{3}{2}}+29{\left(x-9\right)}^{\frac{1}{2}}\right)$

$f\left(13\right)=2\left(8+29\times 2\right)$

$\text{= 132}$