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 know that
$$
\int \frac{1}{t^2+1}dt=\arctan t + C.
$$
Thus, let $u=2t$, then
\begin{align}
\int \frac{6}{4t^2+1}dt &= \int \frac{6}{u^2+1}\frac{1}{2}du\\
&=\int \frac{3}{u^2+1}du\\
&=3\arctan u + C\\
&=3\arctan 2t + C.
\end{align}
Thus $\int \frac{6}{4t^2+1}dt = 6\arctan (2t) + c$, you found, is wrong.
