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Find the value of the constant $c$ for which the integral converges, and evaluate the integral:

$$\int_0^\infty \left(\frac{9x}{x^2+1}-\frac{9c}{2x+1}\right)dx$$

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What are your thoughts so far? – Javier May 12 '13 at 20:04

We have \begin{align} \int_0^a \left(\dfrac{9x}{x^2+1} - \dfrac{9c}{2x+1}\right) dx & = \left. \left(\dfrac92 \log\left(x^2+1 \right) - \dfrac{9c}2 \log(2x+1) \right) \right \vert_0^a\\ & = \dfrac{9}2 \log \left(\dfrac{a^2+1}{(2a+1)^c}\right) \end{align} Given this, as $a \to \infty$, what happens if $c<2$ or $c>2$? Once you get the $c$ from this, use this $c$ and let $a \to \infty$, to get the answer.

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If C is less than 2 then the limit as a approaches infinity is 0? I still dont know how to find the answer for the constant C for which the interval converges? – user77421 May 12 '13 at 20:25
@user77421 Nope. If $c=1$, for instance, what happens? – user17762 May 12 '13 at 20:29

The integrand can be simplified to $$9\frac{2x^2+x-cx^2 -c}{(x^2+1)(2x+1)}.$$ It $c\ne 2$, then for large $x$ the absolute value of the numerator is dominated by the $9(|2-c|x^2)$ term. it follows that for large $x$, the integrand behaves like $\frac{9(|2-c|)}{2x}$, so the integral does not converge.

Thus for convergence we need $c=2$. The calculation of the integral for $c=2$ has been described by user17762.

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so how do i evaluate the integral from 0 to infinity of ((9x/(x^2+1)-(18/(2x+1))dx – user77421 May 12 '13 at 20:38
Like user17762 wrote. Integrate from $0$ to $M$. As written in the answer by user17762, we get $\frac{9}{2}\ln\left(\frac{M^2+1}{(2M+1)^2}\right)$. As $M\to\infty$, the thing inside the $\ln$ approaches $1/4$. – André Nicolas May 12 '13 at 20:44

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