Testing for convergence of this function

Question

For the integral

$$\int_2^\infty \dfrac{x+1}{(x-1)(x^2+x+1)}dx .$$

Can I know if it's convergent or not? If it does can I know how to evaluate it?

I tried to use $u$ substitution but it didn't work.

Answer 1

Hint: Apply partial fractions.

$$\frac{x+1}{(x-1)(x^2+x+1)}=\frac A{x-1}+\frac{Bx+C}{x^2+x+1}$$

$$x+1=(A+B)x^2+(A-B+C)x+(A-C)$$

$$\therefore A=\frac23,B=-\frac23,C=-\frac13$$

Now we know that: $$\int \frac{x+1}{(x-1)(x^2+x+1)}dx=\frac23\int\frac1{x-1}dx-\frac13\int\frac{2x-1}{x^2+x+1}dx\\ =\frac23\ln|x-1|-\frac13\ln(x^2+x+1)=\frac13\ln\frac{(x-1)^2}{x^2+x+1}$$

Therefore,

\begin{align*} \int^\infty_2\frac{x+1}{(x-1)(x^2+x+1)}dx&=\lim_{n\to \infty}\int^n_2\frac{x+1}{(x-1)(x^2+x+1)}dx \\ \\ &=\lim_{n\to\infty}\left(\frac13\ln\frac{(n-1)^2}{n^2+n+1}-\frac13\ln\frac17\right)\\ \\ &=-\frac13\ln\frac17\\ \\ &=\boxed{\dfrac13\ln7} \end{align*}

Answer 2

To see the convergence, just note that, the integrand behaves as

$$ \dfrac{x+1}{(x-1)(x^2+x+1)} \sim_{\infty} \frac{x}{x^3}=\frac{1}{x^2}. $$

Answer 3

Hint: $$0<\int_{2}^{\infty}\frac{x+1}{(x-1)(x^2+x+1)}dx<\int_{2}^{\infty}\frac{x+1}{(x-1)(x^2-1)}dx=\int_{2}^{\infty}\frac{1}{(x-1)^2}dx=1$$

Answer 4

According to The Limit Comparison Test we have $$\lim_{x\to\infty}x^{\mathbf{2}}f(x)=1<\infty$$ so the improper integral converges.