The methods suggested in this thread are mainly specific applications of
Integration by Parts, not part of the VCE Specialist course, but good to know.
A less orthodox method using Specialist knowledge is:
1.
GraphingConsider the definite integral
\; dx)
, where both
Shade in this definite integral. Draw a rectangle from (0,0) to (x2,y2), it can be seen that there are three regions:
Rectangle with corners (0,0) and (x1,y1). Definite integral along the x axis. Definite integral along the y axis.
Thus, the definite integral required can be calculated by:
\; dx + x_1\times y_1 + \int_{y_1}^{y_2} f(y)\; dy = x_2\times y_2)
\; dx = x_2 y_2 - x_1 y_1 - \int_{y_1}^{y_2} f(y)\; dy)
2.
Integral along yHere, the definite integral is evaluated along the y axis. The terminals are now
,\; y_2 = \tan^{-1}(x_2))
and the function is
 = \tan(y))
.
Thus,
\; dy = -\log \left[ \cos(y) \right]_{y_1}^{y_2} = -\log \left[ \frac{\cos(y_2)}{\cos(y_1)} \right] = - \log \left[ \frac{\cos (\tan^{-1}x_2) }{\cos (\tan^{-1} x_1)} \right])
3.
SimplifyingAt this point, draw a right-angled triangle with one of the opposite side length p, adjacent side length 1, thus the hypotenuse is
by Pythagoras' theorem. Thus
 = \frac{p}{1} \implies \theta = \tan^{-1} p)
 = \cos(\theta) = \frac{1}{\sqrt{p^2 + 1}} = (p^2 + 1)^{-1/2})
Thus,
\; dy = - \log \left[ \frac{\cos (\tan^{-1}x_2) }{\cos (\tan^{-1} x_1)} \right] = -\log \left[ \frac{ \left((x_2)^2 + 1\right)^{-1/2} }{\left((x_1)^2 + 1\right)^{-1/2}} \right] = \frac{1}{2} \log \left[ \frac{ (x_2)^2 + 1 }{(x_1)^2 + 1} \right])
Putting everything together:
\; dx = x_2 y_2 - x_1 y_1 - \int_{y_1}^{y_2} f(y)\; dy = x_2 \tan^{-1}(x_2) - x_1 \tan^{-1} (x_1) - \frac{1}{2} \log \left[ \frac{ (x_2)^2 + 1 }{(x_1)^2 + 1} \right])
\; dx = \left(x_2 \tan^{-1}(x_2) - \frac{1}{2}\log \left( (x_2)^2 + 1\right) \right) - \left(x_1 \tan^{-1}(x_1) - \frac{1}{2}\log \left( (x_1)^2 + 1\right) \right))
It can be shown that for integration from negative to positive x values give the same form (break it down into from 0 to x2, and x1 to 0, integration over the negative region is the same as but negative of the positive region due to symmetrical properties).
Thus, by the fundamental theorem of Calculus (part II) that
\; dx = F(b) - F(a))
, it is shown that
 = \int f(x)\; dx = x \tan^{-1} (x) - \frac{1}{2}\log (x^2 + 1) + C)
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