Hi, this is a solution to a question that I couldn't do although I understand the rest I just don't get the following line $\int_{0}^{1}\frac{1}{(x+4)}-\frac{4}{(x+4)^2}dx={[log (x+4)+\frac{4}{x+4}]}^{1}_{0}$ I get how the the $\frac{1}{x+4}$ was integrated but I dont get the fraction next to it Thanks

2. Because $\displaystyle \int{\frac{4}{(x+4)^2}\,dx} = \int{4(x+4)^{-2}\,dx}$

$\displaystyle = \int{4u^{-2}\,du}$ after making the substitution $\displaystyle u = x+4$

$\displaystyle = \frac{4u^{-1}}{-1} + C$

$\displaystyle = -\frac{4}{(x+4)} + C$.

3. Originally Posted by aonin
Hi, this is a solution to a question that I couldn't do although I understand the rest I just don't get the following line $\int_{0}^{1}\frac{1}{(x+4)}-\frac{4}{(x+4)^2}dx={[log (x+4)+\frac{4}{x+4}]}^{1}_{0}$ I get how the the $\frac{1}{x+4}$ was integrated but I dont get the fraction next to it Thanks
$\displaystyle \int -\frac{4}{(x+4)^2} \, dx = -4\int (x+4)^{-2} \, dx = 4(x+4)^{-1} = \frac{4}{x+4}$

in future, post calculus questions in the calculus forum.