How to approach this integration : $\int\frac{4(x^3 - 1)}{4x^5 + 16x^3 + 4x^2 + 1} \; dx$

Question

I just made the following problem to the ways in which i can approach an integral problem.

$$\int\dfrac{4(x^3 - 1)\;dx}{4x^5 + 16x^3 + 4x^2 + 1}$$

I created this problem, such that if it is presented in a certain way would be very easy to integrate. But I then changed the original look of the function.

Now I cant find any way to integrate this, except looking at the original expression.

How can we easily evaluate this integral? It isn't immediately obvious to me many times what to substitute!

Answer 1

Note: As Rene said, there might be an error in the question. There might be $1\over x$ instead of $1$ in the denominator. If there is no error, then ignore the solution below.

We first divide by $x^3$. Subsequently the given integral can be rewritten as: $$\int\dfrac{1-\dfrac{1}{x^3}}{\left(x+\dfrac{1}{2x^2}\right)^2+2^2}$$

Now put $(x+\dfrac{1}{2x^2}) = u$ and $du$ is the numerator.

Answer 2

I guess experience just puts a check list into your head.

First thing I do is check if we aren't integrating $\frac{f'(x)}{f(x) + a}$. Where a is a constant.

Then, I look for standard forms. Maybe there is a $\frac{f'(x)}{f^2(x) +1}$ in there?

Next step is to try factoring terms out, and repeating the process. After that - few steps of partial integration. Are the powers getting lower? Does this feel manageable?

Last step is to plot the integrator and look for some parametrisation.