Evaluate $\int \frac{6x+4}{x^2+4}dx$ Find$$\displaystyle \int \dfrac{6x+4}{x^2+4}dx$$
I'm not really sure where to begin with this one - I know the answer will probably involve an $\arctan$, but I am unsure on how to use $\arctan$ in integrating. A full step by step explanation would be really appreciated.
 A: HINT
Decompose the integral as
$$\int \frac{6x+4}{x^2+4}~\mathrm{d}x = 3\int \frac{2x}{x^2+4}~\mathrm{d}x+4\int\frac{1}{x^2+4}~\mathrm{d}x$$
The first integral on the right will involve $\ln$ and the second will involve $\arctan$.
A: First, you can split the integral up as follows (this is what lab bhattacharjee is referring to):
$$\int\frac{6x+4}{x^2+4}dx=3\int\frac{2x}{x^2+4}dx+4\int\frac{1}{x^2+4}dx.$$
The motivation for doing this is two-fold:


*

*To isolate the function which integrates to something like $\arctan$ - you recognised this.

*To obtain an integrand whose top is the derivative of the bottom - this involves $\ln$.
It is a "standard fact" (which means that it is useful to rememeber) that 
$$\int \frac{1}{1+x^2}dx=\arctan(x) + C.$$ 
You can prove this by using the substitution $x=\tan \theta$.
Now, you are required to find
$$\int\frac{1}{x^2+4}dx$$
What happens if you make the substitution $x=2u$?
A: $$
x^2  + 4 = \frac{1}{4}\left( {\left( {\frac{x}{2}} \right)^2  + 1} \right) \Rightarrow \int {\frac{{dx}}{{x^2  + 4}}}  = 4\int \frac{{dx}}{{\left( {\frac{x}{2}} \right)^2  + 1}}
$$
