Find the value of 
so that the improper integral
![\[ \int_1^{\infty} \left( \frac{x}{2x^2+2C} - \frac{C}{x+1} \right) \, dx \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ee0e5f8588cf036477e0cb3ba01b41f7_l3.png "Rendered by QuickLaTeX.com")
converges, and compute the value of the integral in this case.
First, we compute the value of the constant 
,
![\begin{align*} \int_1^{\infty} \left( \frac{x}{2x^2+2C} - \frac{C}{x+1} \right) \, dx &= \int_1^{\infty} \frac{x^2+x-2C(x^2+C)}{2(x^2+C)(x+1)} \, dx \\[9pt] &= \int_1^{\infty} \frac{(1-2C)x^2 + x - 2C^2}{2(x^2+C)(x+1)} \, dx. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5a7ec63320f43bf72111308a9e341031_l3.png "Rendered by QuickLaTeX.com")
This integral will diverge by comparison with 
if the coefficient of the 
term in the numerator is not equal to zero. Hence, we must have
![\[ 1-2C = 0 \quad \implies \quad C = \frac{1}{2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5438418757ee4f0f8c14fa7a711d430c_l3.png "Rendered by QuickLaTeX.com")
Next, we evaluate the integral with 
,
![\begin{align*} \int_1^{\infty} \left( \frac{x}{2x^2+2C} - \frac{C}{x+1} \right) \, dx &= \int_1^{\infty} \left( \frac{x}{2x^2+1} - \frac{1}{2(x+1)} \right) \, dx \\[9pt] &= \lim_{a \to +\infty} \int_1^a \left( \frac{x}{2x^2+1} - \frac{1}{2(x+1)} \right) \, dx \\[9pt] &= \lim_{a \to +\infty} \left( \frac{1}{4} \int_1^a \frac{4x}{2x^2+1} \, dx - \frac{1}{2} \int_1^a \frac{1}{x+1} \, dx \right) \\[9pt] &= \lim_{a \to +\infty} \left( \frac{1}{4} \log(2x^2+1) - \frac{1}{2} \log |x+1| \right) \Bigr \rvert_1^a \\[9pt] &= \frac{1}{4} \cdot \lim_{a \to +\infty} \left( \log(2a^2+1) - \log 3 - 2 \log (a+1) + 2 \log 2 \right) \\[9pt] &= \frac{1}{4} \cdot \lim_{a\to +\infty} \left( \log \frac{2a^2+1}{(a+1)^2} + \log \frac{4}{3} \right) \\[9pt] &= \frac{1}{4} \cdot \log \frac{8}{3}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8613006a4461d23f7f0e9933c2aa7873_l3.png "Rendered by QuickLaTeX.com")