Finding the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt.$ I was looking for the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt.$ 
My work:
Dividing the $\frac{8t^3 +13}{(t+2)(4t^2+1)}$, I get $2 + \frac{-16t^2 -2t + 9}{4t^3 + 8t^2 + t + 2}$ or $2 + \frac{-16t^2 -2t + 9}{(t+2)(4t^2+1)}$
Using the partial fraction decomposition on the expression $\frac{-16t^2 -2t + 9}{(t+2)(4t^2+1)}$, the partial fractions of $\frac{-16t^2 -2t + 9}{(t+2)(4t^2+1)}$ I get 
is $\frac{-3}{t+2} + \frac{-4t + 6}{4t^2 +1 }$
So...I conclude that....
$$\frac{8t^3 +13}{(t+2)(4t^2+1)} = 2 + \frac{-3}{t+2} + \frac{-4t + 6}{4t^2 +1 } $$
Getting now the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt$.
$$\int \frac{8t^3 +13}{(t+2)(4t^2+1)} dt = \int \left( 2  + \frac{-3}{t+2} + \frac{-4t + 6}{4t^2 +1 } \right) dt $$
$$ = \int 2 dt  + \int \frac{-3}{t+2} dt + \int \frac{-4t + 6}{4t^2 +1 } dt $$
$$ = \int 2 dt  + \int \frac{-3}{t+2} dt + \int \frac{-4t}{4t^2 +1 } dt + \int \frac{6}{4t^2 +1 } dt $$
$$  = 2t  - 3ln(t+2)  - \frac{1}{2}ln(4t^2 + 1) + 6\arctan (2t) + C$$
$$   = 2t  - \left( (ln(t+2)^3)  + ln((4t^2 + 1)^{\frac{1}{2}}) \right) + 6\arctan (2t) + C$$
$$   = 2t  - \left( ln((t+2)^3(4t^2 + 1)^{\frac{1}{2}}) \right) + 6\arctan (2t) + C $$
$$   = 2t  - \frac{1}{2}\left( 2\left( ln((t+2)^3(4t^2 + 1)^{\frac{1}{2}}) \right)\right) + 6\arctan (2t) + C $$
$$   = 2t  - \frac{1}{2}ln((t+2)^6(4t^2 + 1))  + 6\arctan (2t) + C $$
But in my book, it says that the integral of $\frac{8t^3 +13}{(t+2)(4t^2+1)} dt$ is $2t  - \frac{1}{2}ln((t+2)^6(4t^2 + 1))  + 3\arctan (2t) + C$
My answer is so close to that of the book, but I couldn't locate my error. Where did I go wrong?

you should properly pose x=2t but then dx/dt=2 
$I= \int 2 dt  + \int \frac{-3}{t+2} dt + \int \frac{-4t}{4t^2 +1 } dt + \int \frac{6}{ \color{red}{4t^2 +1 }} dt$
$I= \int 2 dt  + \int \frac{-3}{t+2} dt + \int \frac{-4t}{4t^2 +1 } dt + \int \frac{3}{ \color{red}{x^2 +1 }} dx$
$I= \int 2 dt  + \int \frac{-3}{t+2} dt + \int \frac{-4t}{4t^2 +1 } dt + 3\int \frac{1}{ \color{red}{x^2 +1 }} dx$
$I= \int 2 dt  + \int \frac{-3}{t+2} dt + \int \frac{-4t}{4t^2 +1 } dt + 3\color{red}{\arctan(2t)}$
