finding an indefinite integral of a fraction (a) Show that $\frac{4-3x}{(x+2)(x^2+1)}$ can be written in the form ${\frac{A}{x+2} + \frac{1-Bx}{x^2+1}}$ and find the constants $A$ and $B$. 
(b) Hence find $\displaystyle\int\frac{4-3x}{(x+2)(x^2+1)}dx$
For (a) I found that $B=2$ and $A=2$
And I am not quite sure how to integrate. I tried to split them into two $\displaystyle\int\frac{2}{x+2
}dx$ and $\displaystyle\int\frac{1-2x}{x^2+1}dx$ but I don't know how to do after. 

Hint:
Let $u=x+2$ for
\begin{equation}
\int\frac{2}{x+2}dx
\end{equation}
Split the latter integral into two parts
\begin{equation}
\int\frac{1-2x}{x^2+1}dx=\int\frac{1}{x^2+1}dx-\int\frac{2x}{x^2+1}dx
\end{equation}
then let $x=\tan\theta$ also use identity $\sec^2\theta=1+\tan^2\theta$ for
\begin{equation}
\int\frac{1}{x^2+1}dx
\end{equation}
and
let $v=x^2+1$ for
\begin{equation}
\int\frac{2x}{x^2+1}dx
\end{equation}
