Tính tích phân sau: ∫(x^2.3^x/3^x+1)dx

Question

Tính tích phân sau:

\(\int _{-1}^{1}\frac{x^{2} .3^{x} }{3^{x} +1} dx\)

Answer 1

\(\int _{-1}^{1}\frac{x^{2} .3^{x} }{3^{x} +1} dx \)

 Đặt \(x=-t\Rightarrow dx=-dt \)

  Khi đó \(\int _{-1}^{1}\frac{x^{2} .3^{x} }{3^{x} +1} dx =\int _{1}^{-1}\frac{t^{2} .3^{-t} }{3^{-t} +1}  \left(-dt\right)\)

\(=\int _{-1}^{1}\frac{t^{2} }{3^{t} +1} dt =\int _{-1}^{1}\frac{x^{2} }{3^{x} +1} dx \)
\(\Rightarrow \int _{-1}^{1}\frac{x^{2} .3^{x} }{3^{x} +1} dx +\int _{-1}^{1}\frac{x^{2} }{3^{x} +1} dx =\int _{-1}^{1}x^{2} dx=\frac{2}{3} \)

\( \Rightarrow \int _{-1}^{1}\frac{x^{2} .3^{x} }{3^{x} +1} dx =\frac{1}{3}  . \)