`int_0^infty x^3/(x^2+1)^2 dx=`

Substitute `u=x^2+1` `=>` `du=2x dx` `=>` `x dx=(du)/2,` `u_l=0^2+1=1,` `u_u=lim_(x to infty)x^2+1=infty.`

`u_l` and `u_l` denote new lower and upper bounds of integration. We also need to write `x^2` in terms of `u.` From substitution we get `x^2=u-1.` Let us rewrite the integral in a more convenient way before using the above substitution.

`int_0^infty (x^2 x dx)/(x^2+1)^2=`

Now we use the substitution.

`1/2int_1^infty (u-1)/u^2 du=1/2(int_1^infty u/u^2 du-int_1^infty 1/u^2du)=`

`1/2(int_1^infty 1/u du-(- 1/u)|_1^infty)=1/2(ln u+1/u)|_1^infty=`

`1/2(lim_(u to infty)ln u+lim_(u to infty)1/u-ln1-1/1)=1/2(infty+0-0-1)=infty`

As we can see **the integral diverges**.

The image below shows the graph of the function and area under it corresponding to the value of the integral. As we can see the function indeed converges to zero (`x`-axis is the asymptote of the graph of the function) but this convergence is "too slow" to imply the convergence of the integral.