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

Christopher Allers · Accepted Answer

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

Ezekiel Alexander · Answer

undefined To use the integral test to determine convergence or divergence of the given integral, you need to:

1. Evaluate the integral $\int_{1}^{\infty} \frac{x + 1}{x^3 + x^2 + 1} dx$.

2. Determine whether the integral converges or diverges.

Integral tests states that if $f(x)$ is a continuous, positive, and decreasing function for all $x \geq 1$, then the improper integral $\int_{1}^{\infty} f(x) \, dx$ converges if and only if the corresponding series $\sum_{n=1}^{\infty} f(n)$ converges.

You can then use comparison or limit comparison tests to evaluate the integral, if needed, after confirming the necessary conditions are met.