`int_-infty^infty 4/(16+x^2)dx=`

Divide both numerator and denominator by 16.

`int_-infty^infty (1/4)/(1+x^2/16)dx=int_-infty^infty (1/4 dx)/(1+(x/4)^2)=`

Substitute `u=x/4` `=>` `du=1/4 dx.` Bounds of integration remain the same because dividing infinity (or minus infinity) by four gives infinity.

`int_-infty^infty (du)/(1+u^2)=arctan u|_-infty^infty=lim_(u to infty)arctan u-lim_(u to -infty)arctan u=`

`pi/2-(-pi/2)=pi`

**As we can see the integral converges and its value is equal to
`pi.`**

The image below shows the graph of the function and area under it corresponding to the integral. As we can see the graph of the function approaches `x`-axis (function converges to zero) as `x` goes to `+-infty.` We can also see that this convergence seams pretty "fast" which means that the integral should converge which we've already shown by calculating it.

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