Find a value for the constant 
so that the improper integral
![\[ \int_0^{\infty} \left( \frac{1}{\sqrt{1+x^2}} - \frac{C}{x+1} \right) \, dx \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1037cd209233963ecdf4cce95347cfdf_l3.png "Rendered by QuickLaTeX.com")
converges and find the value of the integral in this case.
First, we have
![\begin{align*} \int_0^{\infty} \left( \frac{1}{\sqrt{1+2x^2}} - \frac{C}{x+1} \right) \, dx &= \int_0^{\infty} \frac{x+1-C\sqrt{1+2x^2}}{(x+1)\sqrt{1+2x^2}} \, dx \\[9pt] &= \int_0^{\infty} \frac{x+1-xC\sqrt{2 + \frac{1}{x^2}}}{(x^2+x)\sqrt{2 + \frac{1}{x^2}}} \, dx \\[9pt] &= \int_0^{\infty} \frac{x \left(1 - C \sqrt{2 + \frac{1}{x^2}} \right) + 1}{x^2 \sqrt{2 + \frac{1}{x^2}} + x \sqrt{2 + \frac{1}{x^2}}} \, dx. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-563eb7d0a04094fc166da7e1cb46f484_l3.png "Rendered by QuickLaTeX.com")
For this to converge we must have the coefficient of the 
in the numerator going to 0 as 
(otherwise the integral will converge by limit comparison with 
). Hence,
![\[ \lim_{x \to \infty} \left(1 - C \sqrt{2 + \frac{1}{x^2}} \right) = 1 - \sqrt{2} C \quad \implies \quad C = \frac{\sqrt{2}}{2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b679ecb3488a31abe8a106cd1bf06774_l3.png "Rendered by QuickLaTeX.com")
Now, we need to evaluate the integral. Incomplete.
You already solved earlier (maybe in chapter 8): \int \frac{dx}{\sqrt{1+x^2}}, by substituting t = x + \sqrt{1 + x^2}. This integral can similarly be solved.
No idea how to use LaTeX on this blog, so I’ll just try to be as unambiguous as I can.
The expression on the left has the indefinite integral (arcsinh(sqrt(2)x))/(sqrt(2)), and the ind. integral of the expression on the right is just log(x+1)/sqrt(2), and we simply subtract the second from the first to get the whole integral:
(from 0 to a) integral of (1/sqrt2)*(arcsinh(sqrt(2)x)- log(x+1)) is just
(1/sqrt2)*(arcsinh(sqrt(2)a)- log(a+1)) [since arcsinh(0) = 0, and obv. ln(1) = 0]
Given the relationship between arcsinh and ln, we find arcsinh(sqrt(2)a) = ln(sqrt(2)*a+sqrt(2a^2+1)), so the limit is just (1/sqrt2)*ln({sqrt(2)*a+sqrt(2a^2+1)}/{a+1}),
which is just (3/(2*sqrt2))*ln(2). Not sure why this is off by a factor of 1/2, is it my own error, or the book’s?