How do you use the integral test to determine whether #int (x+1)/(x^3+x^2+1)# converges or diverges from #[1,oo)#?

Question

1680 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-11T08:47:24

The integral is convergent

The integral is

#I=intf(x)dx=int_1^oo((x+1)dx)/(x^3+x^2+1)#

On the interval #I=[1,+oo)#, the function #f(x)# is continuous and #>0#

#I=lim_(x->oo)int_1^t((x+1)dx)/(x^3+x^2+1)#

#f(x)=(x+1)/(x^3+x^2+1)∼_(x->+oo)x/x^3=1/x^2#

By the p-test,

#1/x^2# converges as #p>1#, Riemann's Integral

We conclude that #I=int_1^oo((x+1))dx/(x^3+x^2+1)# converges