Evaluate $\int_{3}^{\infty}\frac{dx}{x^2-x-2}$

Question

$$\int_{3}^{\infty}\frac{dx}{x^2-x-2}=\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}$$

$$\frac{1}{(x-2)(x+1)}=\frac{A}{(x-2)}+\frac{B}{(x+1)}$$

$$1=Ax+A+Bx-2B$$

$$1=(A+B)x+A-2B$$

$A+B=0\iff A=-B$

$-3B=1$

$B=-\frac{1}{3}$, $A=\frac{1}{3}$

$$\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}=\frac{1}{3}\int_{3}^{\infty}\frac{dx}{x-2}-\frac{1}{3}\int_{3}^{\infty}\frac{dx}{(x+1)}=|\frac{1}{3}ln(t-2)-\frac{1}{3}ln(t+1)|_{3}^{\infty}$$

$$=lim_{t\to \infty}(\frac{1}{3}ln(t-2)-\frac{1}{3}ln(t+1))-\frac{1}{3}ln(1)+\frac{1}{3}ln(4)=\infty-\infty-0+\frac{ln(4)}{3}$$

But the answer is $\frac{2ln(2)}{3}$, What have I done wrong?

Answer 1

Observe $\infty-\infty$ is indeterminate form, so you need to get rid from it by writing: $$\lim_{t\to \infty}(\frac{1}{3}\ln(t-2)-\frac{1}{3}\ln(t+1))=\lim_{t\to \infty}\frac{1}{3}\ln{\frac{(t-2)}{(t+1)}}=\frac{1}{3}\ln{1}=0$$

Answer 2

$\frac{\ln(4)}{3} = \frac{\ln(2^2)}{3} = \frac{2\ln(2)}{3}$

And to handle the first two terms, combine the logs:

$\frac{1}{3}\ln(t-2) - \frac{1}{3}\ln(t+1) = \frac{1}{3}\ln(\frac{t-2}{t-1})$

Then taking the limit as $t \rightarrow \infty$ will make it $\frac{1}{3}\ln(1) = 0$.

Answer 3

$$\frac{1}{3}\int_{3}^{+\infty}\left(\frac{1}{x-2}-\frac{1}{x+1}\right)\,dx =\frac{1}{3}\int_{1}^{4}\frac{dz}{z}=\frac{\log 4}{3}=\color{red}{\frac{2}{3}\log 2}\tag{1}$$ since:

$$ \int_{3}^{M}\frac{dx}{x-2}=\int_{1}^{M-2}\frac{dz}{z},\qquad \int_{3}^{M}\frac{dx}{x+1}=\int_{4}^{M+1}\frac{dz}{z}\tag{2} $$ and: $$ \left|\int_{M-2}^{M+1}\frac{dz}{z}\right|\leq \frac{1}{M-2}\int_{M-2}^{M+1}dz = \frac{3}{M-2}\to 0.\tag{3}$$