Integral Calculation

What's the integral of:
[tex] \frac{ {x}^{2} + 1}{ {e}^{2} + 1 } [/tex]
from -1 to 1 ?​

I know that it should be equal to 4/3, but how do we reach to that conclusion?

Respuesta :

Answer:

[tex]\frac{4}{3(e^2+1)}[/tex]

Step-by-step explanation:

We want to evaluate:

[tex]\int\limits^1_{-1} {\frac{x^2+1}{e^2+1} } \, dx[/tex]

This is the same as:

[tex]\frac{2}{e^2+1} \int\limits^1_{0} {x^2+1} \, dx[/tex]

We integrate to obtain:

[tex]\frac{2}{e^2+1} (\frac{x^3}{3}+x)|_0^1[/tex]

We evaluate to obtain:

[tex]\frac{2}{e^2+1} (\frac{1^3}{3}+1)=\frac{4}{3(e^2+1)}[/tex]