Tính tích phân sau: ∫dx/(e^x+1)(x^2+1)

Question

Tính tích phân sau:

\(\int _{-1}^{1}\frac{dx}{(e^{x} +1)(x^{2} +1)} \)

Answer 1

\(I=\int _{-1}^{1}\frac{dx}{(e^{x} +1)(x^{2} +1)}\)  

 Đặt \(u=-x\Rightarrow du=-dx\)

 Đổi cận: \(x=-1\Rightarrow u=1;x=1\Rightarrow u=-1\)
\(\Rightarrow I=\int _{-1}^{1}\frac{dx}{(e^{x} +1)(x^{2} +1)}  =\int _{-1}^{1}\frac{e^{u} du}{(e^{u} +1)(u^{2} +1)} \) 

\(\Rightarrow 2I=\int _{-1}^{1}\frac{dx}{(e^{x} +1)(x^{2} +1)}  +\int _{-1}^{1}\frac{e^{x} dx}{(e^{x} +1)(x^{2} +1)}\)

\( =\int _{-1}^{1}\frac{dx}{x^{2} +1}   \)
Đặt \(x=\tan t\Rightarrow dx=\frac{1}{{\rm cos}^{2} t} dt=(\tan ^{2} t+1)dt\)

 Đổi cận: \(x=-1\Rightarrow t=-\frac{\pi }{4} ;x=1\Rightarrow u=\frac{\pi }{4}\) 

 Ta có \(2I=\int _{-\frac{\pi }{4} }^{\frac{\pi }{4} }\frac{(1+\tan ^{2} t)dt}{1+\tan ^{2} t} \)

\(=\int _{-\frac{\pi }{4} }^{\frac{\pi }{4} }dt=\left. t\right|_{-\frac{\pi }{4} }^{\frac{\pi }{4} } =\frac{\pi }{2}  \Rightarrow I= \frac{\pi }{4} .\)