Computing $ \int \frac{{\rm d} x}{x^3 + 12} $

Question

$$ \int \frac{{\rm d} x}{x^3 + 12} $$

This is a question I came up with and have not been able to solve. I graphed this function and it is as in the picture attached. Some of the people who I have talked to have labeled it as an "impossible integral." Is it true or IS there actually a way to do this? I have recently started learning integration, but am really intrigued and curious.

Your help and explanation will mean a lot. Thank you so much!

Answer 1

First decompose the integrand into partial fractions $$\frac{1}{x^3+a^3}=\frac{1}{(x+a)(x^2-ax+a^2)}=\frac{E}{x+a}+\frac{Fx+G}{x^2-ax+c}$$ where $a=12^{1/3}.$You shouln't have any trouble finding $E,F,G$ or in inegating the first fraction as a natural logarithm and the second fraction in terms of a natural logarithm and an arctan.

Answer 2

Decompose the integrand into three manageable components below

$$\frac{1}{x^3+a^3} =\frac{a(x+a) -x^2 +(x^2-ax+a^2)}{2a^2(x^3+a^3)}\\ =\frac1{2a^2}\left(\frac{a}{x^2-ax+a^2}-\frac{x^2}{x^3+a}+\frac{1}{x+a}\right)$$

Answer 3

here's a solution without decomposition.

substitute -> x=cbrt(12y), as mentioned by a previous answerer, this results in in the aforementioned integral (12^2/3)*integral of (1/y^3+1), then according to sum of cubes we have the following result -> (y+1)(y^2-y+1)=y^3+1, hence y=tan(u), dy=sec2(u) as in secant squared of u or sec u times sec u, then we have the following tanu +1 times sec2(u)-tanu, that is after some algebraic manipulation, this becomes cos^2u/1-sinucosu.then u may continue with division