Biết $\Large \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{{{x}^{2}}dx}{{{(

Biết $\Large \int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{{{x}^{2}}dx}{{{(x\sin x+\cos x)}^{2}}}=-\dfrac{a\pi }{b+c\pi \sqrt{3}}+d\sqrt{3}}$, với $\Large a,b,c,d$ là các số nguyên dương. Tính $\Large P=a+b+c+d$ 

• A.

  $\Large P=9$

• B.

  $\Large P=10$

• C.

  $\Large P=8$

• D.

  $\Large P=7$

Đặt $\Large I=\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{{{x}^{2}}dx}{{{\left( x\sin x+\cos x \right)}^{2}}}}$

$\Large =\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{{{x}^{2}}({{\tan }^{2}}x+1)}{{{(x\tan x+1)}^{2}}}dx}$

$\Large =\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{{{x}^{2}}{{\tan }^{2}}x+2x\tan x+1-2x\tan x-2+1+{{x}^{2}}}{{{(x\tan x+1)}^{2}}}dx}$

$\Large =\int\limits_{0}^{\dfrac{\pi }{3}}{\left[ 1-\dfrac{2}{x\tan x+1}+\dfrac{{{x}^{2}}+1}{{{(x\tan x+1)}^{2}}} \right]dx}$

Xét tích phân $\Large K=\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{{{x}^{2}}+1}{{{(x\tan x+1)}^{2}}}dx}$

Đặt $\Large \left\{ \begin{align}  & u={{x}^{2}}+1\Rightarrow du=2xdx \\  & dv=\dfrac{1}{{{(x\tan x+1)}^{2}}}dx\Rightarrow v=\dfrac{\tan x}{x\tan x+1} \\ \end{align} \right.$

$\Large K=({{x}^{2}}+1).\dfrac{\tan x}{x\tan x+1}\left| \begin{align}  & \dfrac{\pi }{3} \\  & 0 \\ \end{align} \right.$ $\Large -\int\limits_{0}^{\dfrac{\pi }{3}}{\dfrac{2x\tan x}{x\tan x+1}dx=}$ $\Large ({{x}^{2}}+1).\dfrac{\tan x}{x\tan x+1}\left| \begin{align}  & \dfrac{\pi }{3} \\  & 0 \\ \end{align} \right.-\int\limits_{0}^{\dfrac{\pi }{3}}{\left( 2-\dfrac{2}{x\tan x+1} \right)dx}$

$\Large \Rightarrow I=-\dfrac{\pi }{3}+\left( \dfrac{{{\pi }^{2}}}{9}+1 \right)\dfrac{\sqrt{3}}{\dfrac{\pi }{3}.\sqrt{3}+1}=\dfrac{-\pi +3\sqrt{3}}{\pi \sqrt{3}+3}=-\dfrac{4\pi }{3+\pi \sqrt{3}}+\sqrt{3}$

$\Large \Rightarrow a=4;b=3;c=1;d=1\Rightarrow P=9$