How do I show $\int_0^\infty \frac{x}{(x+1)^2}dx$ diverges I did $$\int_0^\infty \frac{x}{(x+1)^2}dx=\int_0^\infty \frac{1}{(x+2+\frac1x)}dx\le\int_0^\infty \frac{1}{x}dx$$ but $\int_0^\infty \frac{1}{x}dx$ diverges. So my logic fails.
 A: $\begin{eqnarray}\int_0^\infty\frac{x}{(x+1)^2}\mathrm{d}x&=&\lim_{t\to\infty}\int_0^t\frac{x}{(x+1)^2}\mathrm{d}x\\&=&\left(\lim_{t\to\infty}\frac{1}{t+1}+\log(t+1)\right)-1\\&=&\infty\end{eqnarray}$
A: By the limit comparison test:
$$\frac{\frac x{(1+x)^2}}{\frac1x}=\frac{x^2}{(1+x)^2}\xrightarrow[x\to\infty]{}1$$
and thus our integral converges iff $\;\int\limits_1^\infty\frac{dx}x\;$ converges. But it's easy to show directly that this last integral diverges, so does ours.
Observe that for $\;\int\limits_0^1\frac x{(1+x)^2}dx\;$ there is no problem, since this a standard Riemann itnegral of a continue function in $\;[0,1]\;$
A: You can split the fraction $\dfrac{x}{(1+x)^2}$ into two parts: $\dfrac{x}{(1+x)^2} = \dfrac{1}{1+x} - \dfrac{1}{(1+x)^2}$. The first integral $\displaystyle \int_{0}^\infty \dfrac{1}{1+x}dx$ diverges while the second integral $\displaystyle \int_{0}^\infty \dfrac{1}{(1+x)^2}dx$ converges, thus the given integral diverges.
A: Let $c$ be the upper limit of the integral and take the limit as $c$ goes to infinity.
$$I = \int_0^\infty \frac{x}{(x+1)^2}dx = \lim_{c \rightarrow \infty} \int_0^c \frac{x}{(x+1)^2}dx.$$
Then compute the definite integral as normal;
$$I = \lim_{c \rightarrow \infty}\int_0^c \frac{x}{(x+1)^2} = \lim_{c\rightarrow\infty}\left[\log(x + 1) - \frac{1}{x+1}\right]_0^c = 1 + \lim_{c \rightarrow \infty} \left(\log(c+1) - \frac{1}{c+1}\right).$$
The logarithm diverges as its argument gets arbitrarily large and the $\frac{1}{c+1}$ term vanishes. Hence,
$$\lim_{c \rightarrow \infty} \int_0^c \frac{x}{(x+1)^2}dx = \infty$$
and we conclude that $I$ diverges.
A: We have for $x\ge 1$ and arbitrary $\varepsilon > 0$
$\frac{x}{(x+1)^2} > \frac{x+\varepsilon}{(x+\varepsilon+1)^2}$.
This gives
$\int_{0}^{\infty}\frac{x}{(x+1)^2}\space\text{d}x\\
=\lim_{n\to\infty}\int_{0}^{n+1}\frac{x}{(x+1)^2}\space\text{d}x\\
=\int_{0}^{1}\frac{x}{(x+1)^2}\space\text{d}x
+\lim_{n\to\infty}\int_{1}^{n+1}\frac{x}{(x+1)^2}\space\text{d}x\\
=(log(2)-1/2)
+\lim_{n\to\infty}\sum_{k=1}^{n}\int_{k}^{k+1}\frac{x}{(x+1)^2}\space\text{d}x\\
> (log(2)-1/2)
+\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k+1}{(k+2)^2} = \infty.$
Where we crudely estimated the integral from below and used the harmonic series for the last step.
A: You can also show it by evaluating the integral:
$$\int_{0}^{\infty}\frac{x}{(x+1)^2}\space\text{d}x=\lim_{n\to\infty}\int_{0}^{n}\frac{x}{(x+1)^2}\space\text{d}x=$$
$$\lim_{n\to\infty}\int_{0}^{n}\left[\frac{1}{x+1}-\frac{1}{(x+1)^2}\right]\space\text{d}x=\lim_{n\to\infty}\left[\int_{0}^{n}\frac{1}{x+1}\space\text{d}x-\int_{0}^{n}\frac{1}{(x+1)^2}\space\text{d}x\right]=$$

Substitute $u=x+1$ and $\text{d}u=\text{d}x$.
This gives a new lower bound $u=0+1=1$ and upper bound $u=n+1$:

$$\lim_{n\to\infty}\left[\int_{1}^{n+1}\frac{1}{u}\space\text{d}u-\int_{1}^{n+1}\frac{1}{u^2}\space\text{d}u\right]=\lim_{n\to\infty}\left[\left[\ln(u)\right]_{1}^{n+1}+\left[\frac{1}{u}\right]_{1}^{n+1}\right]=$$
$$\lim_{n\to\infty}\left[\ln(1)-\ln(n+1)+\frac{1}{1}-\frac{1}{n+1}\right]=\lim_{n\to\infty}\left[0-\ln(n+1)+1-\frac{1}{n+1}\right]=$$
$$\lim_{n\to\infty}\left[1-\frac{1}{1+n}-\ln(1+n)\right]\space\to-\infty$$
