Improper Integral Question: $ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx $ I have to test the convergence of the integral : 
$$ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx $$
Please suggest. Also, have to show that the value of the integral is zero ?

You want to find
$$ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx $$
First, we find a primitive of the integrand, this is, we need
$$ \int\frac{x\log x}{(1+x^2)^2} dx =F(x)$$
Use integration by parts with $u = \log x$ and $dv = \dfrac{x}{(1+x^2)^2}$.
This makes $du = dx/x$ and $v=-\dfrac{1}{2(1+x^2)}$, from where
$$ \int\frac{x\log x}{(1+x^2)^2} dx = -\dfrac{\log x}{2(1+x^2)}+\int \dfrac{dx}{2x(1+x^2)}$$
To find the last integral, use partial fractions:
$$\frac{1}{{2x\left( {1 + {x^2}} \right)}} = \frac{1}{2}\left( {\frac{1}{x} - \frac{x}{{1 + {x^2}}}} \right)$$
This means
$$\int {\frac{{dx}}{{2x\left( {1 + {x^2}} \right)}}}  = \frac{1}{2}\log x - \frac{1}{4}\log \left( {1 + {x^2}} \right)$$
So the primitive we're looking for is
$$ \int\frac{x\log x}{(1+x^2)^2} dx = -\dfrac{\log x}{2(1+x^2)}+\frac{1}{2}\log x - \frac{1}{4}\log \left( {1 + {x^2}} \right)$$
Writing this more tidily, 
$$\int {\frac{{x\log x}}{{{{(1 + {x^2})}^2}}}} dx =  - \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]$$
Now we need to evaluate those expressions at $x \to 0$, and $x \to \infty$.
$$\int\limits_0^\infty  {\frac{{x\log x}}{{{{(1 + {x^2})}^2}}}dx}  =  - \mathop {\lim }\limits_{x \to \infty } \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right] + \mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]$$
You can convince yourself pretty easily that
$$\mathop {\lim }\limits_{x \to \infty } \frac{{\log x}}{{1 + {x^2}}} = \mathop {\lim }\limits_{x \to \infty } \log \frac{{\sqrt {1 + {x^2}} }}{x} = 0$$
so all it is needed is to find
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]$$
which is an indeterminate form $\infty -\infty$.
This might be really tedious to solve, so I will use something that might be frowned upon:
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \log \frac{{\sqrt {1 + {x^2}} }}{x}} \right]=$$
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\frac{{\log x}}{{1 + {x^2}}} + \frac{1}{2}\log \left( {\frac{1}{{{x^2}}} + 1} \right)} \right] = $$
Here's the catch: for $x \to 0$
$$\eqalign{
  & 1 + {x^2} \sim 1  \cr 
  & \frac{1}{{{x^2}}} + 1 \sim \frac{1}{{{x^2}}} \cr} $$
so the limit can be simplified to 
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left[ {\log x + \frac{1}{2}\log \left( {\frac{1}{{{x^2}}}} \right)} \right]=$$
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{2}\left( {\log x - \log x} \right) = 0$$
So
$$ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx =0$$
This is why maybe using subtitutions is more recommended.
