Calculate Limit - Is There a Special Rule?

[Petrus]

Hold on. Not so fast.
You're skipping a couple of steps.

You wanted to calculate
$$\int_{2n - \frac 1 2}^{6n + \frac 1 2} \frac{n}{k^2+n^2} dk$$

What you now have, is (note the last back substitution to x):
$$\int \frac{3}{x^2+3^2}dx = \int \frac{1}{u^2+1}du = \arctan u + C = \arctan \frac x 3 + C$$

First you need to generalize that to the actual integral that contains n.

Then you can (carefully!) substitute the boundaries.

And only then can you calculate the limit for n to infinity.

If I got this right we got $$\arctan \frac{k}{n}+C$$

 

[Opalg]

Hmm... when I do that and subsitute our limits to $$[\arctan{u}]_{\frac{4n-1}{6}}^{\frac{12n+1}{6}}$$ we get and then calculate limit we get $$\arctan(infinity)-arctan(infinty)$$ the answer shall be $$\arctan(6)-\arctan(2)$$ I am doing something wrong?

$$\arctan(6)-\arctan(2)$$ is the correct answer. I would do this problem like this: write $$\sum_{k=2n}^{6n} \frac n{k^2+n^2}$$ as $$\sum_{k=2n}^{6n} \frac1n\,\frac 1{\bigl(\frac kn\bigr)^2+1}$$, then recognise this as a Riemann sum for the integral $$\int_2^6\frac1{x^2+1}\,dx.$$