integration of $\int_{1}^{\infty } \,\left(\frac{2x^{2}+bx\text{+}a}{x(2x+a)} -1\right) \, dx=1$ i need help for this problem;
Find values of a and b $$\int_{1}^{\infty} \left( \frac{2x^{2}+bx+a}{x(2x+a)} -1\right) \, dx=1$$
I very appreciate your comments and suggestions.
 A: Hint: Use long division on the integrand. You should get $$\int_1^\infty \frac{-a+b-2}{a+2 x}+\frac 1x+1-1$$
$$=\int_1^\infty\frac{-a+b-2}{a+2 x}+\int_1^\infty\frac 1x$$
$$=\lim_{t\to\infty}(-a+b-2)\int_1^t \frac{1}{a+2x}+\int_1^t \frac 1x$$
Can you take it from there?
A: For 
\begin{align}
I = \int_{1}^{\infty} \left( \frac{2 x^2 + b x + a}{x(2x+a)} - 1 \right) \, dx = 1
\end{align}
then it is seen that
\begin{align}
I &= \lim_{p \to \infty} \, \int_{1}^{p} \left( \frac{b-a}{2x+a} + \frac{a}{x(2x+a)} \right) \, dx = \lim_{p \to \infty} \, \int_{1}^{p} \left( \frac{b-a-2}{2x+a} + \frac{1}{x} \right) \, dx \\
&= \lim_{p \to \infty} \, \left[ \frac{b-a-2}{2} \, \ln(2x+a) + \ln(x) \right]_{1}^{p} \\
&= \lim_{p \to \infty} \ln\left[ p \, \left(\frac{2p+a}{2+a} \right)^{\frac{b-a-2}{2}} \right] = \ln(e) = 1
\end{align}
for which
\begin{align}
\lim_{p \to \infty} \left\{ p \, \left(\frac{2p+a}{2+a} \right)^{\frac{b-a-2}{2}} \right\} = e. 
\end{align}
The limit is evaluated as follows: 
\begin{align}
e &= \lim_{p \to \infty} \left\{ p \left(\frac{2p+a}{2+a}\right)^{\frac{b-a-2}{2}} \right\} \\
&= \left(\frac{2}{a+2}\right)^{\frac{b-a-2}{2}} \, \lim_{p \to \infty} \left\{ p^{\frac{b-a}{2}} \, \left( 1 + \frac{a}{2p}\right)^{\frac{b-a-2}{2}} \right\} = \left(\frac{2}{a+2}\right)^{\frac{b-a-2}{2}} \, \lim_{p \to \infty} \left\{ p^{\frac{b-a}{2}} \right\} 
\end{align}
If $b=a$ then the limit is reduced to $1$. This yields
\begin{align}
e = \frac{a + 2}{2}
\end{align}
leading to $a = b = 2(e - 1)$.
