Evaluate the integral $$\int_{0}^{1}\frac{3x+4}{x^3-3x-4}\mathrm{d}x$$
What should be the approach for evaluating this integral? No substitution seems to be working and the denominator also doesn't have good roots so partial fraction method would be too tedious.
Best Answer
The computation is manageable.
The denominator can be factored as
$$x^3-3x-4=(x-a)(x^2+ax+c)$$
where $a$ is the real root, and
$$c=a^2-3=\dfrac4a.$$
Then if you decompose the function as
$$\frac{2A}{x-a}-\frac{A(2x+a)}{x^2+ax+c}+\frac B{x^2+ax+c},$$
every term is "easy" to integrate. By identification,
$$\begin{cases}3Aa+B=-3,\\3A(a^2-2)-aB=-4.\end{cases}$$