Cho $\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{xdx}{1-{{\sin }^{2}}x}}=\dfrac{\pi }{a}-\ln b+\ln \sqrt{2};a,b\in \mathbb{N}*.$ Giá trị $a+3b$ bằng

Phương pháp:
Áp dụng phương pháp tích phân từng phần.
Cách giải:
Ta có $\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{xdx}{1-{{\sin }^{2}}x}}=\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{xdx}{{{\cos }^{2}}x}}$
Đặt $\left\{ \begin{aligned}
& x=u \\
& dv=\dfrac{dx}{{{\cos }^{2}}x} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& dx=du \\
& v=\tan x \\
\end{aligned} \right.$
$\Rightarrow I=x\tan x\left| \begin{aligned}
& \dfrac{\pi }{4} \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\dfrac{\pi }{4}-\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x}{\cos x}dx}$
$=\dfrac{\pi }{4}+\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{d\left( \cos x \right)}{\cos x}}=\dfrac{\pi }{4}+\ln \left| \cos x \right|\left| \begin{aligned}
& \dfrac{\pi }{4} \\
& 0 \\
\end{aligned} \right.$
$=\dfrac{\pi }{4}+\ln \dfrac{\sqrt{2}}{2}=\dfrac{\pi }{4}+\ln \sqrt{2}-\ln 2$
$\Rightarrow a=4,b=2$
Vậy $a+3b=10.$