How to show that these integrals don't converge I'm trying to solve these two improper integrals $$\int_{1}^{\infty} \dfrac{x+4}{2x^2 +x -3}\, dx$$ and $$\int_{-\infty}^{-1} \dfrac{x+4}{2x^2 +x -3}\, dx = 0$$
I evaluated the indefinite integral and then I tried to solve the definite integral using substitution, but it didn't really work out for me. I checked out WolframAlpha, but there's no step-by-step solution for these integrals. Could someone please show me how to solve them step-by-step, as after hours of attempts I keep failing to do so.
 A: As written by Andrei
$$f(x)=\frac{x+4}{2x^2+x-3}$$
$$=\frac{x+4}{(x-1)(2x+3)}$$
$$\sim \frac{5}{5(x-1)} \;\; (x\to 1)$$
but
$$\int_1^2\frac{dx}{x-1} \text{ is divergent}$$
thus
$$\int_1^2f(x)dx \text{ is divergent }$$
and so does $\int_1^{+\infty}f(x)dx$.
A: You need to do partial fraction decomposition:
$$\frac{x+4}{2x^2+x-3}=\frac{x+4}{(2x+3)(x-1)}=\frac A{2x+3}+\frac B{x-1}$$
Then $$A(x-1)+B(2x+3)=x+4$$or $$A+2B=1\\-A+3B=4$$
Add the last two equations, and you get $5B=5$ or $B=1$, and therefore $A=-1$.
Then your integral becomes much simpler. Can you take it from here?
A: Since $1$ is a pole, let first simplify by making the change $u=x-1$
$$\int_1^\infty \frac{x+4}{2x^2+x-3}\,dx=\int_0^\infty\frac {u+5}{u(2u+5)}\,dx=\int_0^\infty\frac{du}u-\int_0^\infty\frac{du}{2u+5}$$
The first integral has an issue in $0$ (i.e. $\ln(u)\to-\infty$), the second one is continuous there (the second integral is so congergent in $0$) so the issue in $0$ cannot be removed by summing and the overall integral is divergent.
But even in the case the integral would have started at $x=2$, we still have an issue at infinity:
Indeed for $u\gg 1$ we have $\begin{cases}u+5>u\\2u+5<3u\end{cases}\implies \dfrac{u+5}{u(2u+5)}>\dfrac 1{3u}$
Which is divergent at infinity (i.e. $\ln(u)\to+\infty$), so the integral is divergent again.
For the last part of the exercise, you similarly have a pole in $-\frac 32$ and the same issue at minus infinity.
