Find a value for the constant 
so that the improper integral
![\[ \int_2^{\infty} \left( \frac{Cx}{x^2+1} - \frac{1}{2x+1} \right) \, dx \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b1911e5cec78c312bb0b08438d2b4991_l3.png "Rendered by QuickLaTeX.com")
converges. Find the value of the integral for this value of 
.
First, we compute the value of ,
![\begin{align*} \int_2^{\infty} \left( \frac{Cx}{x^2+1} - \frac{1}{2x+1} \right) \, dx &= \int_2^{\infty} \left( \frac{Cx(2x+1) - (x^2+1)}{(x^2+1)(2x+1)} \right) \, dx \\[9pt] &= \int_2^{\infty} \frac{(2C-1)x^2 + Cx - 1}{(2x+1)(x^2+1)} \, dx. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5d7be46a2ce9d9ac6fd776021a6645bb_l3.png "Rendered by QuickLaTeX.com")
Since this integral will diverge by limit comparison with 
for any nonzero coefficient of the 
term in the numerator we must have
![\[ 2C - 1 = 0 \quad \implies \quad C = \frac{1}{2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-74da7fcde775cdc40023ee9801106450_l3.png "Rendered by QuickLaTeX.com")
Next, we evaluate the integral
![\begin{align*} \int_2^{\infty} \left( \frac{Cx}{x^2+1} - \frac{1}{2x+1} \right) \, dx &= \int_2^{\infty} \left( \frac{\frac{1}{2}x}{x^2+1} - \frac{1}{2x+1} \right) \, dx \\[9pt] &= \lim_{a \to +\infty} \left( \frac{1}{2} \int_2^a \frac{x}{x^2+1} \, dx - \int_2^a \frac{1}{2x+1} \, dx \right) \\[9pt] &= \lim_{a \to +\infty} \left( \frac{1}{4} \log (x^2+1) - \frac{1}{2} \log|2x+1| \right) \Bigr \rvert_2^a \\[9pt] &= \frac{1}{4} \cdot \lim_{a \to +\infty} \left( \log(a^2+1) - \log 5 - 2 \log (2a+1) + 2 \log 5 \right) \\[9pt] &= \frac{1}{4} \cdot \lim_{a \to +\infty} \left( \log \frac{a^2+1}{(2a+1)^2} + \log 5 \right) \\[9pt] &= \frac{1}{4} \cdot \left( \log \frac{1}{4} + \log 5 \right) \\[9pt] &= \frac{1}{4} \log \frac{5}{4}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7ce06357575b65787af120bead3923e0_l3.png "Rendered by QuickLaTeX.com")