`int_0^oo x^3/(x^2+1)^2 dx` Determine whether the integral diverges or converges. Evaluate the integral if it converges.

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`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.  

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