Find values for real numbers 
and 
so that the following improper integral equation holds:
![\[ \int_1^{\infty} \left( \frac{2x^2 + bx + a}{x(2x+a)} - 1 \right) \, dx = 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-583e60d7681e0e5c8f185aa846ef85ee_l3.png "Rendered by QuickLaTeX.com")
First, we have
![\begin{align*} \int_1^{\infty} \left( \frac{2x^2 + bx + a}{x(2x+a)} - 1 \right) \, dx &= \int_1^{\infty} \frac{2x^2 + bx + a - 2x^2 - ax}{2x^2+ax} \, dx \\[9pt] &= \int_1^{\infty} \frac{(b-a)x + a}{2x^2 + ax} \, dx. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e78e92f2e3ba4868487cbc7548304020_l3.png "Rendered by QuickLaTeX.com")
In order for this to converge we must have the coefficient of 
in the numerator equal to 0; hence, we must have 
. Now, to solve the integral equation we have
![\begin{align*} \int_1^{\infty} \frac{a}{2x^2 + ax} \,dx &= \int_1^{\infty} \frac{a}{x(2x+a)} \, dx \\[9pt] &= \int_1^{\infty} \left( \frac{1}{x} - \frac{2}{2x+a} \right) \, dx &(\text{partial fractions})\\[9pt] &= \lim_{c \to \infty} \int_1^c \left( \frac{1}{x} - \frac{2}{2x+a} \right) \, dx \\[9pt] &= \lim_{c \to \infty} \left( \log |x| - \log |2x+a| \right)\Bigr \rvert_1^c \\[9pt] &= \lim_{c \to \infty} \left( \log \left| \frac{x}{2x+a} \right| \right) \Bigr \rvert_1^c \\[9pt] &= \lim_{c \to \infty} \left( \log \left( \frac{c}{2c+a} \right) - \log \left( \frac{1}{2+a} \right) \right) \\[9pt] &= \log \frac{1}{2} - \log \frac{1}{2+a} \\[9pt] &= \log \frac{2+a}{2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e83eaf9fbee5b6349f51530a51e88785_l3.png "Rendered by QuickLaTeX.com")
But, by assumption this integral equals 1, so we have
![\begin{align*} \log \frac{2+a}{2} = 1 && \implies && \frac{2+a}{2} &= e \\[9pt] && \implies && 2+a &= 2e \\[9pt] && \implies && a & =2e - 2. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-91331abd291e5af33bbbb91e68531b68_l3.png "Rendered by QuickLaTeX.com")
Since 
we then have
![\[ a = b = 2e - 2. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-04280c9888edc59c00ccf01f819e1625_l3.png "Rendered by QuickLaTeX.com")
I believe that you used the comparison test to state that a=b. If I’m right how can you assure that the analyzed function is >0?
In the second integral, from fifth to sixth row, why did you remove the module? If you keep the module until the end, other solution gonna appear. The other solution is a= -2e-2.