# Integrate $\int\frac{1}{x^3+1}dx$

The problem is, as stated:

$$\int\frac{1}{x^3+1}dx$$

I tried using substitution: $t^3 = x^3 + 1$ but didn't get far with that. I also tried setting: $t = x^3 + 1$, with no luck again.

I tried partial decomposition but I didn't know how to integrate $$\int\frac{1}{x^2-x+1}$$ and I kept getting that term when expanding $x^3 + 1$

Any help would be much appreciated.

• @ Luka Duranovic use $x^2-x+1=(x-\frac12)^2+\frac34$
– Minz
Apr 7 '18 at 11:16

Hint for the last integral

$$I=\int\frac{1}{x^2-x+1}=\int\frac{1}{(x-1/2)^2+3/4}=\int\frac{du}{u^2+3/4}$$

• yw @LukaDuranovic ...then substitute to get the arctan function Apr 7 '18 at 11:19
• Right. You can always complete the square to get arctan for something like that. Apr 7 '18 at 11:19
• I smell some kind of fishy in my second method of the answer I posted can you please give it a look whether it is good or is there any mistake Apr 7 '18 at 16:52

$$\int\frac {1}{x^2-x+1} dx=\int\frac {1}{(x-\frac{1}{2})^2 +\frac{3}{4}}dx$$ And that take $x-\frac{1}{2}=t$. Always use $ax^2+bx+c=a(x+\frac{b}{2a}) ^2 - \frac{b^2-4ac}{4a}$

Method 1

$$\int \frac {dx}{x^2-x+1}=\frac {4}{3}\int \frac {dx}{1+\frac {4(x-1/2)^2}{3}}$$

Put $u=x-1/2$

Hence $$\frac 43\int \frac {dx}{1+\frac {4(x-1/2)^2}{3}} =\frac 43\int \frac {du}{1+\frac {4u^2}{3}}$$

And note that $$\int \frac {dx}{1+x^2}=\arctan x$$

I hope you can take it from here