Biết I = landa x^2-1 / x căn bậc hai x^2 -4 dx= pi / 2a + b căn bậc hai 3 + c ( a,b,c thuộc Z). Khi đó, P= -a + b+ c bằng

Biết $I={\int }_{2\sqrt{2}}^4\frac{x^2-1}{x\sqrt{x^2-4}}dx=\frac{\pi }{2a}+b\sqrt{3}+c\left(a,b,c\in {\mathbb{Z}}^{*}\right)$. Khi đó, $P=-\alpha +b+c$ bằng

Trả lời

Cách 1. Ta có

$\begin{array}{l}={\int }_{2\sqrt{2}}^4\frac{x^2-1}{x\sqrt{x^2-4}}dx={\int }_{2\sqrt{2}}^4\left(\frac{x}{\sqrt{x^2-4}}-\frac{1}{x\sqrt{x^2-4}}\right)dx\\ ={\int }_{2\sqrt{2}}^4\frac{x}{\sqrt{x^2-4}}dx-{\int }_{2\sqrt{2}}^4\frac{1}{x\sqrt{x^2-4}}dx\\ ={\int }_{2\sqrt{2}}^4\frac{x}{\sqrt{x^2-4}}dx-\frac{\pi }{24}\end{array}$

Đặt $t=\sqrt{x^2-4}\Rightarrow t^2=x^2-4\Rightarrow tdt=xdx$

Với $x=2\sqrt{2}\Rightarrow t=2,x=4\Rightarrow t=2\sqrt{3}$

$\begin{array}{l}\Rightarrow I={\int }_2^{2\sqrt{3}}dt-\frac{\pi }{24}={\left.t\right|}_2^{2\sqrt{3}}-\frac{\pi }{24}=2\sqrt{3}-2-\frac{\pi }{24}\\ \Rightarrow a=-12,b=2,c=-2\Rightarrow P=12\end{array}$

Cách 2. Đặt $x=\frac{2}{\cos t}$. Ta có $\left\{\begin{array}{l}dx=\frac{-2\sin t}{{\cos }^{2}t}dt\\ \frac{2}{4}=\frac{2}{\cos \frac{\pi }{3}}\\ \frac{2}{2\sqrt{2}}=\frac{2}{\cos \frac{\pi }{3}}\end{array}\right.$

$\Rightarrow I={\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{\left(\frac{4}{{\cos }^{2}t}-1\right)}{\frac{2}{\cos t}\sqrt{\frac{4}{{\cos }^{2}t}-4}}\frac{-2\sin tdt}{{\cos }^{2}t}=\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{{\cos }^{2}t-4}{{\cos }^{2}t}dt=\frac{\pi }{24}-2\sqrt{3}+2$

$\Rightarrow a=-12,b=2,c=-2\Rightarrow P=12$

Chọn D