Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 38

Differentiate $\displaystyle y = \frac{x^3 - 1}{x^2 + 1 } + 4x^3$
By applying long division, we get



Thus,
$\displaystyle y = \left[ x + \frac{-x - 1}{x^2 + 1} \right] + 4x^3$

Then, by taking the derivative, we obtain

$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} (x) + \frac{(x^2 + 1) \cdot \frac{d}{dx} ( - x - 1) - ( - x - 1) \cdot \frac{d}{dx} (x^2 + 1 )}{(x^2 + 1)^2} + \frac{d}{dx} (4x^3)\\
\\
y' &= 1 + \frac{(x^2 + 1)(-1) - (-x-1)(2x)}{(x^2 +1)^2} + 12x^2\\
\\
y' &= 1 + \left[ \frac{-x^2 -1 + 2^2 + 2x}{(x^2 +1)^2} \right] + 12x^2\\
\\
y' &= 1 + \left[ \frac{x^2 + 2x - 1}{(x^2 + 1)^2} \right] + 12x^2 \\
\\
y' &= \frac{(x^2 + 1)^2 + x^2 + 2x - 1+ 12x^2 (x^2 + 1)^2}{(x^2 + 1)^2}\\
\\
y' &= \frac{x^4 + 2x^2 + 1 + x^2 + 2x - 1 + 12 x^6 + 24x^4 + 12x^2}{(x^2 + 1)^2}\\
\\
y' &= \frac{12x^6 + 25x^4 + 15x^2 + 2x}{(x^2 + 1)^2}
\end{aligned}
\end{equation}
$