Using $\arctan(x)$ for integrating the following function We want to find $\int \dfrac{x}{x^2 + 16} dx$
My method was as follows:
Rewrite it to: $\dfrac{\frac{1}{16x}}{{(\frac{1}{4x}})^2 +1}$
Take $u = \dfrac{1}{4}x$. We then have $ \int4u^2 \cdot \dfrac{1}{u^2+1} du = \dfrac{4}{3}u^3 \cdot \arctan(u) + c$.
However, this is not correct. What have I done wrong?
Also, my textbook uses the much easier method with $\ln$. I am however questioning how you could judge by 'looks' of the function whether to use the $\arctan$ or the $\ln$. 
 A: You need to be systematic and use substitution correctly. As to how to know what method to use, after a while you will become efffective at it. At this time the priority is correct execution.
At the end there is the assertion 
$$\int 4u^2 \cdot \frac{1}{1+u^2}=\frac{4}{3}u^3 \arctan u+C.$$
This is incorrect, the integral of a product is not the product of the integrals.
There are also errors earlier, though it is not easy to identify them clearly since no steps are given.
The first algebraic step is correct, although not helpful. Then there is the substitution $u=\frac{1}{4} x$.  Perhaps that is a typo, and $\frac{1}{4x}$ was intended. 
Whichever is the case, if we are going to make a substitution, we need to compute explicitly $dx$ as $g(u)\,du$ for the appropriate $g$. There is no indication that this process was followed. We are just told that the integral is equal to $\int 4u^2\cdot \frac{1}{u^2+1}\,du$. This is not true under the proposed substitution, or any variant of it.
Remark: For $\int \frac{x}{x^2+16}\,dx$, you will after a while see that the derivative of $x^2 +16$ is (almost) sitting on top. So soon the response try $u=x^2+16$ will become semi-automatic. 
