Answer
$$\int \frac{16 t+4 t^{3} -t^{2}}{t^{2}+4} dt =2 t^{2}-t+2 \tan^{-1}\frac{t}{2}+C\\$$
Work Step by Step
Given $$\int \frac{16 t+4 t^{3} -t^{2}}{t^{2}+4} dt $$
So, we have
\begin{aligned}
I&=\int \frac{16 t+4 t^{3} -t^{2}}{t^{2}+4} d t\\
&=\int \frac{4 t\left(4+t^{2}\right)-t^{2}}{\left(t^{2}+4\right)} d t\\
&=\int 4 t d t-\int \frac{t^{2}}{\left(t^{2}+4\right)} d t\\
&=4 \cdot \frac{t^{2}}{2}-\int \frac{\left(t^{2}+4\right)-4}{\left(t^{2}+4\right)} d t\\
&=2 t^{2}-\int d t+4 \int \frac{d t}{t^{2}+4}\\
&\text{since}\int \frac{d t}{z^{2}+a^2}= \frac{1}{a} \tan^{-1}\frac{z}{a}, \ \text{so, we get}\\
I&=2 t^{2}-t+\frac{4}{2} \tan^{-1}\frac{t}{2}+C\\
&=2 t^{2}-t+2 \tan^{-1}\frac{t}{2}+C\\
\end{aligned}
