Limit $\lim_{n\to\infty}\int _{2n}^{3n}\frac{x+3}{x^2+1}\,dx?$ How to evaluate
$$\lim_{n\to\infty}\int _{2n}^{3n}\frac{x+3}{x^2+1}\,dx?$$
The answer in the book is $\ln\frac a2$ where $a$ is a constant.
Attempt:
$\lim_{n\to\infty }\int_{2n}^{3n}\:\frac{x+3}{x^2+1}dx =\lim_{n\to\infty}[\frac12\ln(x^2+1)+3\arctan x]^{3n}_{2n}=\lim_{n\to\infty}[\frac12\ln(9n^2+1)+3\arctan3n-\frac12\ln(4n^2+1)-3\arctan2n]=\infty
+\frac{3\pi}2-\infty-\frac{3\pi}2$
and no $\ln\frac a2$. Can someone explain to me how to evaluate this limit?
Thanks!!
 A: Since $n \to \infty$ you can asymptotically approximate $$\frac{x+3}{x^2+1} \sim \frac{1}{x}$$
Hence the limit becomes
$$\lim_{n \to \infty} \int_{2n}^{3n} \frac{x+3}{x^2+1} \ \mathrm{d} x =
\lim_{n \to \infty} \int_{2n}^{3n} \frac{1}{x} \ \mathrm{d} x = \lim_{n \to \infty} \ln (3n/2n) = \ln (3/2)$$
A: Start from the last limit expression:
$$\lim_{n\to\infty}\left(\frac12\ln(9n^2+1)+3\arctan3n-\frac12\ln(4n^2+1)-3\arctan2n\right)$$
The arctangent terms both tend to $\frac{3\pi}2$ as you have noticed, so they cancel in the limit and can be removed:
$$=\lim_{n\to\infty}\left(\frac12\ln(9n^2+1)-\frac12\ln(4n^2+1)\right)$$
Now combine the two logarithms into one:
$$=\lim_{n\to\infty}\frac12\ln\frac{9n^2+1}{4n^2+1}$$
Because $\lim_{n\to\infty}\frac{9n^2+1}{4n^2+1}=\frac94$, we can replace it and obtain the desired form:
$$=\lim_{n\to\infty}\frac12\ln\frac94$$
$$=\ln\frac32$$
Hence $a=3$.
A: Hint: write for the indefinite integral
$$\frac{1}{2}\int\frac{2x}{x^2+1}dx$$ and $$3\int\frac{1}{x^2+1}dx$$
A: Well, for the integral:
$$\mathscr{I}_\text{n}:=\int_{2\text{n}}^{3\text{n}}\frac{3+x}{1+x^2}\space\text{d}x=\int_{2\text{n}}^{3\text{n}}\frac{3}{1+x^2}\space\text{d}x+\int_{2\text{n}}^{3\text{n}}\frac{x}{1+x^2}\space\text{d}x\tag1$$


*

*Substitute $\text{u}=1+x^2$:
$$\int_{2\text{n}}^{3\text{n}}\frac{x}{1+x^2}\space\text{d}x=\frac{1}{2}\int_{1+\left(2\text{n}\right)^2}^{1+\left(3\text{n}\right)^2}\frac{1}{\text{u}}\space\text{d}\text{u}=\ln\left|1+\left(3\text{n}\right)^2\right|-\ln\left|1+\left(2\text{n}\right)^2\right|\tag2$$

*$$\int_{2\text{n}}^{3\text{n}}\frac{3}{1+x^2}\space\text{d}x=3\int_{2\text{n}}^{3\text{n}}\frac{1}{1+x^2}\space\text{d}x=3\cdot\left(\arctan\left(3\text{n}\right)-\arctan\left(2\text{n}\right)\right)\tag3$$


So, we get:
$$\mathscr{I}_\text{n}=\ln\left|1+\left(3\text{n}\right)^2\right|-\ln\left|1+\left(2\text{n}\right)^2\right|+3\cdot\left(\arctan\left(3\text{n}\right)-\arctan\left(2\text{n}\right)\right)\tag4$$
Now, we know that $\text{n}>0$:
$$\mathscr{I}_\text{n}=\ln\left(\frac{1+\left(3\text{n}\right)^2}{1+\left(2\text{n}\right)^2}\right)+3\cdot\left(\arctan\left(3\text{n}\right)-\arctan\left(2\text{n}\right)\right)\tag5$$
So, for the limit:
$$\text{L}:=\lim_{\text{n}\to\infty}\space\mathscr{I}_\text{n}=\lim_{\text{n}\to\infty}\left\{\ln\left(\frac{1+\left(3\text{n}\right)^2}{1+\left(2\text{n}\right)^2}\right)+3\cdot\left(\arctan\left(3\text{n}\right)-\arctan\left(2\text{n}\right)\right)\right\}=$$
$$\lim_{\text{n}\to\infty}\ln\left(\frac{1+\left(3\text{n}\right)^2}{1+\left(2\text{n}\right)^2}\right)=\ln\left(\lim_{\text{n}\to\infty}\frac{18\text{n}}{8\text{n}}\right)=\frac{1}{2}\ln\left(\frac{9}{4}\right)=\ln\left(\frac{3}{4}\right)\tag6$$
A: Letting $u=\frac{x}{n}$, then
\begin{eqnarray}
&&\lim_{n\to\infty}\int _{2n}^{3n}\frac{x+3}{x^2+1}\,dx\\
&=&\lim_{n\to\infty}\int _{2}^{3}\frac{nu+3}{n^2u^2+1}\,ndu\\
&=&\int_2^3\frac1udu\\
&=&\ln\frac32.
\end{eqnarray}
