Chủ đề này có 1 trả lời

[rabbit]

$\int_{-1}^{1}\frac{1}{\left ( x^2+1 \right )\left ( 4^x+1 \right )}dx$

[quocdu89]

$\int_{-1}^{1}\frac{1}{\left ( x^2+1 \right )\left ( 4^x+1 \right )}dx$

$A=\int_{-1}^{1}\dfrac{1}{\left ( x^2+1 \right )\left ( 4^x+1 \right )}dx$

Đặt $t=-x$, sau khi đổi cận ta có:

$A=-\int_{1}^{-1}\frac{1}{\left ( t^2+1 \right )\left ( 4^{-t}+1 \right )}dt=\int_{-1}^{1}\frac{4^{t}}{\left ( t^2+1 \right )\left ( 4^{t}+1 \right )}dt$

Hay $A=\int_{-1}^{1}\frac{4^{x}}{\left ( x^2+1 \right )\left ( 4^{x}+1 \right )}dx=B$

(Vì $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(t)dt$)

Khi đó:

$A+B=\int_{-1}^{1}\frac{1+4^{x}}{\left ( x^2+1 \right )\left ( 4^{x}+1 \right )}dx=\int_{-1}^{1}\frac{1}{\left ( x^2+1 \right )}dx$

Đặt: $x=tan(k)\Rightarrow dx=(1+tan^{2}k)dk$ , sau khi đổi cận ta có:

$A+B=\int_{-1}^{1}\frac{1}{\left ( x^2+1 \right )}dx=\int_{-\frac{\pi}{4} }^{\frac{\pi}{4}}\frac{1}{\left ( tan^2k+1 \right )}.(tan^2k+1)dk=\frac{\pi}{2}$

Vậy $\left\{\begin{matrix}A=B\\A+B=\dfrac{\pi}{2} \end{matrix}\right.\Rightarrow A=\dfrac{\pi}{4}$