Cho hàm số y = f(x) liên tục trên đoạn [-2; 2] và $\large 2f(x)+3f(-x)

Cho hàm số y = f(x) liên tục trên đoạn [-2; 2] và $\large 2f(x)+3f(-x)=\dfrac{1}{x^2+4}, \forall x\in [-2; 2]$.Tính $\large I=\int_{-2}^2 f(x)dx$

• A. $\large I=\dfrac{\pi}{10}$
• B. $\large I=-\dfrac{\pi}{10}$
• C. $\large I=-\dfrac{\pi}{20}$
• D. $\large I=\dfrac{\pi}{20}$

+ Ta có: $\large 2f(x)+3f(-x)=\dfrac{1}{x^2+4}, \forall x\in[-2; 2]$, suy ra $\large 2\int_{-2}^2f(x)dx+3\int_{-2}^2f(-x)dx=\int_{-2}^2\dfrac{1}{x^2+4}dx$

$\large \Leftrightarrow 2\int_{-2}^2f(x)dx-3\int_{-2}^2f(-x)d(-x)=\int_{-2}^2\dfrac{1}{x^2+4}dx$

$\large \Leftrightarrow \int_{-2}^2f(x)dx-3\int_{2}^{-2}f(x)dx=\int_{-2}^2\dfrac{1}{x^2+4}dx \Leftrightarrow 2\int_{-2}^2f(x)dx+3\int_{-2}^2f(x)dx=\int_{-2}^2\dfrac{1}{x^2+4}dx$ 

$\large \Leftrightarrow 5\int_{-2}^2f(x)dx=\int_{-2}^2\dfrac{1}{x^2+4}dx\Leftrightarrow I=\dfrac{1}{5}\int_{-2}^2\dfrac{1}{x^2+4}dx$ 

+ Tính: $\large A=\int_{-2}^2\dfrac{1}{x^2+4}dx$ 

Đặt: $\large x=2\tan t\Rightarrow dx=2(1+\tan^2t)dt, t\in \left(-\dfrac{\pi}{2}; \dfrac{\pi}{2}\right)$

Đổi cận: $\large \left\{\begin{align}& x=-2\Rightarrow t=-\dfrac{\pi}{4}\\& x=2\Rightarrow t=\dfrac{\pi}{4}\\\end{align}\right.$. Khi đó, ta có

$\large A=\int_{-\pi/4}^{\pi/4}\dfrac{1}{4\tan^2t+4}.2(1+\tan^2t)dt=\dfrac{1}{2}\int_{-\pi/4}^{\pi/4}dt=\dfrac{\pi}{4}$

Vậy: $\large \int_{-2}^2 f(x)dx=\dfrac{\pi}{20}$