1. ## Integration

Use the substitution x = 1/u to find the value of

$\int_{\frac{1}{2}}^{2} \frac {lnx}{1+x^2}dx$

2. Hi

$x=\frac{1}{u} \implies \mathrm{d}x=-\frac{\mathrm{d}u}{u^2}$

Substitute this in the integral, it should give something interesting.

3. $\displaystyle x=\frac{1}{u}\Rightarrow dx=-\frac{1}{u^2}du$
$\displaystyle x=\frac{1}{2}\Rightarrow t=2, \ x=2\Rightarrow t=\frac{1}{2}$.
So, $\displaystyle I=\int_{\frac{1}{2}}^2\frac{\ln x}{1+x^2}dx=\int_2^{\frac{1}{2}}\frac{\ln u}{1+u^2}du=-\int_{\frac{1}{2}}^2\frac{\ln u}{1+u^2}du=-I\Rightarrow I=0$

4. Note how the wording is just a little strange. "to find" isn't the usual task, is it?

Do the substitution. What happens?

Note the symmetry of the substitution with the limits of integration.

Hint: How many Numbers, R, can you think of where R = -R?