<s>[INST] Evaluating a Discontinuous Integral How do you evaluate this integral? $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$
My work:
The expression $x^2 - 2x - 3$ is discontinuous at $x = 3$ in the interval $x = 0$ to $x = 4$, so I got to integrate the expression like this:
$$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ is equal to
$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3} $$.
Then recall that in evaluating a discontinuous integrand, the integral is defined by the relations $$\int_a ^b f(x) = lim_{x->b^-} \int_a ^x f(x) dx $$
if $x = b$  is the discontinuous point or $$\int_a ^b f(x) = lim_{x->a^+} \int_x ^b f(x) dx $$
if $x = a$  is the discontinuous point.
With that in mind:
$$\int _0 ^3\frac{dx}{x^2 - 2x - 3} + \int _3 ^4\frac{dx}{x^2 - 2x - 3} $$
$$\int _0 ^3\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)} + \int _3 ^4\frac{dx}{(x^2 - 2x + 1) + (-3)(-1)} $$
$$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} + \int _3 ^4\frac{dx}{(x-1)^2  -(2)^2} $$
Remembering that $\int \frac{du}{u^2 +a^2} = \frac{1}{a} \arctan \left( \frac{u}{a}\right) + C, $
we get:
$$\int _0 ^3\frac{dx}{(x-1)^2 -(2)^2} +\int _3 ^4\frac{dx}{(x-1)^2 -(2)^2}$$
equals
$$lim_{x->3^+} \int_0 ^x \frac{dx}{(x-1)^2 -(2)^2} + lim_{x->3^-} \int_x ^4 \frac{dx}{(x-1)^2 -(2)^2} $$
equals
$$lim_{x->3^+} \left( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_3 ^x + lim_{x->3^-} \left(\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right)|_x ^4$$
equals
$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{3-2}{2}\right) \right ) + \left ( \frac{1}{2} \arctan \left( \frac{4-2}{2}\right)-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$
equals
$$\left ( \frac{1}{2} \arctan \left( \frac{x-2}{2}\right)- 0.2318 \right ) + \left ( \frac{\pi}{8}-\frac{1}{2} \arctan \left( \frac{x-2}{2}\right) \right )$$
which makes  $$\int _0 ^4\frac{dx}{x^2 - 2x - 3} = 0.1609$$
But in my book, it said there is no value of $\int _0 ^4\frac{dx}{x^2 - 2x - 3}.$
How do you prove that the integral $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ has no value?
 [/INST] In fact, you should use the expression for integral of the function $\dfrac{1}{x^2-a^2}$:
$$
\int_3^4 \frac{dx}{(x-1)^2-2^2} =  \int_2^3 \frac{dx}{x^2-2^2} = \frac{1}{4}\lim_{x\to 2^-}\left (\log\frac{1}{5} -\log\frac{x-2}{x+2}\right ).
$$
The above limit does not exist.
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