Integral question - $\int\frac{(4-x)\,dx}{x^2+4x+8}$ Integral question - $$\int\frac{(4-x)\,dx}{x^2+4x+8}$$
To solve it I need to bring the numerator to be the derivative of the dominator right?
I need to do the trick that not change the integral any ideas?
$$\frac{1}{2}\int\frac{(2x+4) \, dx}{x^2+4x+8}$$
Thanks

We have
$$4-x=-\frac{1}{2}(2x+4)+6$$
and 
$$x^2+4x+8=(x+2)^2+4=4((\frac{1}{2}(x+2))^2+1)=4(u^2+1)$$
hence
$$\int\frac{(4-x)\,dx}{x^2+4x+8}=-\frac{1}{2}\int\frac{(2x+4)}{x^2+4x+8}dx+6\int\frac{dx}{x^2+4x+8}\\=-\frac{1}{2}\log|2x^2+4x+8|+{3}\int\frac{du}{u^2+1}=-\frac{1}{2}\log|2x^2+4x+8|+3\arctan(\frac{x+2}{2})+C$$
