Hey dude, here's the gist:
y = log<sub>e</sub>x ---------> x = e<sup>y</sup>
I'll take that question from the cambridge 18 b) 28 as an example.
(Find ∫ lnx dx between 1 and a)
At x=a , y=lna . Draw the perpendiculars from the x and y axis to (a,lna) and think in terms of a rectangle. The area of the rectangle is a.log<sub>e</sub>a and the area you want to get rid of is equal to the integral of e<sup>y</sup> between 0 and lna.
∫ e<sup>y</sup> dy between 0 and lna
= e<sup>(log<sub>e</sub>a)</sup> - e<sup>0</sup>
= a - 1
Hence the area of the area you're looking for is:
a.log<sub>e</sub>a - (a -1)
= a.log<sub>e</sub>a -a + 1
I hope that gives you the general idea.