$$I = \int_1^2 \left(e^{1-\frac{1}{(x-1)^2}} +1\right) + \left(1 +\frac{1}{\sqrt{1- \log(x-1)} }\right)\mathrm{d}x$$.
From what I have read it has to do with forming a triangle, but I do not know where this comes from. If someone can please explain the evaluation of this it would be greatly appreciated. Thanks!
The correct form of the integral (see the video) is: $$ I = \int_1^2 \left( e^{1-\frac1{(x-1)^2}}+ 1\right) + \left( 1+ \frac1 {\sqrt{1-\log(x-1)}} \right)\,dx. $$ This equals $$ \int_1^2 (f(x) + f^{-1}(x))\,dx $$ where $f(x):=\exp\left({1-\frac1{(x-1)^2}}\right)+1$. Check that $f$ is increasing and $f(1)= 1$ and $f(2)= 2$, so the value of the integral is $$ 2f(2)-1f(1)=3 $$ using the formula $$\int_a^b f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = bf(b)-af(a).$$ This is much easier to see geometrically: Borrowing a picture from this answer, you can visualize the integral as the area of an elbow-shaped region:
which in this case can be decomposed into three squares of unit area.
