Integral convergence $ \int_1^ \infty \frac{x+2}{(x+1)^3}dx$

I'm trying to solve this problem:

Determine the convergence of the integral:

$$ \int_1^ \infty \frac{x+2}{(x+1)^3}dx $$

So I know that the integral exist (by using wolfram) so it converges, what I tried is to find a function $f(x)$ s.t. $\frac{x+2}{(x+1)^3}< f(x)$, $\forall x \geq 0$. and the integral of this $f(x)$ should converge in the same interval.

One of the attempts I think can work is:

$$ \frac{x+2}{(x+1)^3}<\frac{x+2}{(x+1)^2}$$

Because if I write $(x+1)$ the numerator is bigger then it will be harder to work the integral...

Probably a function kind of a "p-serie" it's what I'm looking because I know those functions converge

Any suggestions to find that $f(x)$?

Here's a tip to get you started. Try:

$$\int \frac{x+2}{(x+1)^3} dx= \int \frac{x+1}{(x+1)^3} + \frac{1}{(x+1)^3} dx$$

$(x+2)/(x+1)^3\le 1/x^2$ for $x>1$. Indeed $(x+2)x^2=x^3+2x<x^3+3x+3x^2+1$ if $x>0$. So you can take $f(x)=1/x^2$. The integral of $1/x^2$ from $1$ to $\infty$ converges unlike $\int_1^\infty \frac{x+2}{(x+1)^2} dx$.

$\color{blue}{{\rm Thm}}$: Let

• $f$ continuous and positive function on $[a;+\infty[$.
• $g$ continuous and positive function on $[a;+\infty[$.

Therefore,

$$f\underset{+\infty}{\sim}g \implies \int_{a}^{+\infty}f(x){\rm d}x<+\infty \iff \int_{a}^{+\infty} g(x){\rm d}x<+\infty.$$

Proof of convergence to $\displaystyle \int_{1}^{+\infty} \frac{x+2}{(x+1)^{3}}{\rm d}x$:

Let, $$\displaystyle f: x\mapsto \frac{x+2}{(x+1)^{3}}, \quad I=\int_{1}^{+\infty}f(x){\rm d}x,\quad g:x \mapsto \frac{1}{x^{2}}.$$

• The function $g$ is continuous and positive on $[1;+\infty[$.
• The function $f$ is continuous and positive on $[1;+\infty[$ then we have a impropriety in $+\infty$.

Since $f\underset{+\infty}{\sim}g$ because $$\lim_{x\to +\infty} \frac{f(x)}{g(x)}=\lim_{x\to +\infty} \frac{\frac{x+2}{(x+1)^{3}}}{\frac{1}{x^{2}}}=\lim_{x\to +\infty} \frac{x^{2}(x+2)}{(x+1)^{3}}=1.$$ Therefore we have $I$ converges (by $\color{blue}{{\rm Thm}}$) because it's well-know that $\displaystyle \int_{1}^{+\infty}\frac{1}{x^{2}}{\rm d}x$ converges.