Integration by parts: Integrate the expression x.ln(x) between 1 and 2.

Let $ denote the integral symbol, as I am limited here by my keyboard.

Recall the formula for integration by parts:

$ u.(dv/dx) dx = uv - $ u(dv/dx) dx

So to find $^pi₀x.sin(x) dx, we must allocate x and sin(x) to u and (dv/dx). Integrating sin(x) is simpler than integrating x, so let x = u and sin(x) = (dv/dx). 

From the formula, we also need to know v and (du/dx), which we can find by integrating (dv/dx) and differetiating u. So:

(du/dx) = (d/dx)x = x

v=$(dv/dx)dx = $sin(x) dx = -cos(x)

So now we have:

$^pi₀x.sin(x) dx = [x.-cos(x)]^pi₀ - $^pi₀-cos(x) dx

= (pi.1 - 0) + $^pi₀cos(x) dx

=pi + [sin(x)]^pi₀ 

=pi + (0 -0)

= pi.