Back to the class
In class we did the following integral which works very nicely: $\displaystyle\int \dfrac{4x^3+3}{x^4+3x} \mathrm{d}x$ - it is straightforward by taking $u=x^4+3x$. Someone in class asked how the following similar looking integral would work: $\displaystyle\int \dfrac{4x^3+4}{x^4+3x} \mathrm{d}x$. Firstly, I would rewrite the numerator as $4x^3+3+1$ and then split the integral into two:
$$(*) \hspace{35pt} \displaystyle\int \dfrac{4x^3+3+1}{x^4+3x} \mathrm{d}x = \displaystyle\int \dfrac{4x^3+3}{x^4+3x} \mathrm{d}x + \displaystyle\int \dfrac{1}{x^4+3x} \mathrm{d}x.$$
Here the first of the two integrals on the right is the one we can already solve by straightforward $u$-substitution. The second one seems like a problem. The easiest way to solve it is to notice the following algebraic fact:
$$\dfrac{1}{x^4+3x} = \dfrac{1}{3x} - \dfrac{x^2}{3(x^3+3)}.$$
To see this is true, start with the right-hand-side and get a common denominator:
$$\begin{array}{ll}
\dfrac{1}{3x} - \dfrac{x^2}{3(x^3+3)} &= \dfrac{(x^3+3) - x(x^2)}{3(x)(x^3+3)} \\
&=\dfrac{3}{3(x^4+3x)} \\
&= \dfrac{1}{x^4+3x},
\end{array}$$
as was to be shown. After this, it is simple to find $\displaystyle\int \dfrac{1}{x^4+3x} \mathrm{d}x$:
$$\begin{array}{ll}
\displaystyle\int \dfrac{1}{x^4+3x} \mathrm{d}x &= \displaystyle\int \dfrac{1}{3x} - \dfrac{x^2}{3(x^3+3)} \mathrm{d}x \\
&= \displaystyle\int \dfrac{1}{3x} \mathrm{d}x - \displaystyle\int \dfrac{x^2}{3(x^3+3)} \mathrm{d}x \\
&= \dfrac{1}{3} \displaystyle\int \dfrac{1}{x} \mathrm{d}x - \dfrac{1}{3} \displaystyle\int \dfrac{x^2}{x^3+3} \mathrm{d}x \\
\end{array}$$
At this point the first integral is simple and the second integral can be evaluated by taking $u=x^3+3$ yielding $\mathrm{d}u = 3x^2 \mathrm{d}x$, hence $\dfrac{1}{3} \mathrm{d}u = x^2 \mathrm{d} x$. Therefore we have shown
$$\displaystyle\int \dfrac{1}{x^4+3x} \mathrm{d}x = \dfrac{1}{3} \ln(|x|) - \dfrac{1}{9} \ln(|x^3+3|) + C.$$
Now this can be used in $(*)$ to finally solve the original integral (using the result we got in class).
Note: you may wonder "how on earth would I find that 'algebraic fact'?" The answer is partial fraction decomposition --- a topic that we will cover in calc 2 in Section 8.5.
