sum_(n=1)^oo n/(n^4+2n^2+1) Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

sum_(n=1)^oon/(n^4+2n^2+1)
The integral test is applicable if f is positive, continuous and decreasing function on the infinite interval [k,oo) where k>=1 and a_n=f(x) . Then the series converges or diverges if and only if the improper integral int_k^oof(x)dx converges or diverges.
For the given series a_n=n/(n^4+2n^2+1)
Consider f(x)=x/(x^4+2x^2+1)
f(x)=x/(x^2+1)^2
From the attached graph of the function, we can see that the function is continuous, positive and decreasing on the interval [1,oo)
We can also determine whether f(x) is decreasing by finding the derivative f'(x) such that f'(x)<0 for x>=1 .
Apply the quotient rule to find the derivative,
f'(x)=((x^2+1)^2d/dx(x)-xd/dx(x^2+1)^2)/(x^2+1)^4
f'(x)=((x^2+1)^2-x(2(x^2+1)2x))/(x^2+1)^4
f'(x)=((x^2+1)(x^2+1-4x^2))/(x^2+1)^4
f'(x)=(-3x^2+1)/(x^2+1)^3
f'(x)=-(3x^2-1)/(x^2+1)^3<0
Since the function satisfies the conditions for the integral test, we can apply the integral test.
Now let's determine the convergence or divergence of the improper integral as follows:
int_1^oox/(x^2+1)^2dx=lim_(b->oo)int_1^bx/(x^2+1)^2dx
Let's first evaluate the indefinite integral intx/(x^2+1)^2dx
Apply integral substitution:u=x^2+1
=>du=2xdx
intx/(x^2+1)^2dx=int1/(u^2)(du)/2
=1/2int1/u^2du
Apply the power rule,
=1/2(u^(-2+1)/(-2+1))
=-1/(2u)
Substitute back u=(x^2+1)
=-1/(2(x^2+1))+C  where C is a constant
Now int_1^oox/(x^2+1)^2dx=lim_(b->oo)[-1/(2(x^2+1))]_1^b
=lim_(b->oo)-1/2[1/(b^2+1)-1/(1^2+1)]
=-1/2[0-1/2]
=1/4
Since the integral int_1^oox/(x^4+2x^2+1)dx converges, we conclude from the integral test that the series sum_(n=1)^oon/(n^4+2n^2+1) converges.