Question

Evaluate the integral or show it is divergent. I'm uploading an image of how I did this. I'm just not confident with any of this I would like someone to tell me if I approached this correctly.
Thanks!

Answer 1

\[\int_{-1}^{1}\frac{dx}{x^2-x}\]
First mistake: \(\dfrac{1}{x^2-2x}\not=\dfrac{1}{x^2}-\dfrac{1}{2x}\).

Instead, the easiest method would be to decompose the integrand into partial fractions:
\[\frac{1}{x^2-2x}=\frac{1}{x(x-2)}=\frac{A}{x}+\frac{B}{x-2}\\
1=A(x-2)+Bx\]
which gives the system
\[\begin{cases}A+B=0\\-2A=1\end{cases}~~\Rightarrow~~A=-\frac{1}{2},~B=\frac{1}{2}\]
So, the integral becomes
\[-\frac{1}{2}\int_{-1}^1\frac{dx}{x}+\frac{1}{2}\int_{-1}^1\frac{dx}{x-2}\]
The second integrand is defined over [-1,1], so you can carry on with the computation as usual, but before you do that...

The first must be treated as an improper integral due to the singularity at 0:
\[\lim_{b\to0^-}\int_{-1}^b\frac{dx}{x}+\lim_{a\to0^+}\int_a^1\frac{dx}{x}\\
\lim_{b\to0^-}\bigg[\ln |x|\bigg]_{-1}^b+\lim_{a\to0^+}\bigg[\ln |x|\bigg]_a^1\\
\lim_{b\to0^-}\ln |b|-\ln|-1|+\ln |1|-\lim_{a\to0^+}\ln|a|\\
\lim_{b\to0^+}\ln b-0+0-\lim_{a\to0^+}\ln a\\
\lim_{b\to0^+}\ln b-\lim_{a\to0^+}\ln a\]
Neither limit converges to a finite number, so the first integral diverges. Hence the original integral diverges.

Answer 2

Thank you so much!!

Answer 3

yw