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

Question

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

Attempt:

I added and subtracted $x^3$ in the numerator but after a little solving I can't get through.

Answer 1

$$\begin{align} \int\frac{1}{1+x^3}\,dx &=\int\frac{1-x^2+x^2}{1+x^3}\,dx\\ &=\int\frac{1-x^2}{1+x^3}\,dx+\int\frac{x^2}{1+x^3}\,dx\\ &=\int\frac{1-x}{1-x+x^2}\,dx+\frac13\ln\left(1+x^3\right)+C\\ &=\int\frac{1-(u+1/2)}{1-(u+1/2)+(u+1/2)^2}\,du+\frac13\ln\left(1+x^3\right)+C\\ &=\int\frac{1/2-u}{3/4+u^2}\,du+\frac13\ln\left(1+x^3\right)+C\\ &=\frac12\int\frac{1}{3/4+u^2}\,du-\int\frac{u}{3/4+u^2}\,du+\frac13\ln\left(1+x^3\right)+C\\ &=\frac12\frac{1}{\sqrt{3/4}}\arctan\left(\frac{u}{\sqrt{3/4}}\right)-\frac12\ln\left(3/4+u^2\right)+\frac13\ln\left(1+x^3\right)+C\\ &=\frac{1}{\sqrt{3}}\arctan\left(\frac{x-1/2}{\sqrt{3/4}}\right)-\frac12\ln\left(1-x+x^2\right)+\frac13\ln\left(1+x^3\right)+C\\ \end{align}$$

Answer 2

Does this help?

$\dfrac{1}{x^3+1} = \dfrac{1}{(x+1)(x^2-x+1)} = \dfrac13\left(\dfrac{1}{x+1} - \dfrac{x-2}{x^2-x+1}\right)$

Can you integrate it from there?

The integral of the second part is a little tricky. Note that:

$\dfrac13\left(\dfrac{x-2}{x^2-x+1}\right) = \dfrac16\left(\dfrac{2x-4}{x^2-x+1}\right)= \dfrac16\left(\dfrac{2x-1}{x^2-x+1} - \dfrac{3}{x^2-x+1}\right)$

$=\dfrac16\left(\dfrac{2x-1}{x^2-x+1} - \dfrac{3}{\left(x-\frac12\right)^2 + \frac34}\right)$

That first integral is a $u$-substitution, and the second uses the arctangent function.

Answer 3

I thought it might be instructive to show how we can evaluate the more general integral $I_n(x)$ given by

$$I_n(x)=\int \frac{1}{1+x^n}\,dx$$

for $n\ge 1$

In THIS ANSWER, I provided a step-by-step presentation that showed

$$\bbox[5px,border:2px solid #C0A000]{\int\frac{1}{x^n+1}dx=-\frac1n\sum_{k=1}^n\left(\frac12 x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)+C'} \tag 1 $$

where $x_{kr}$ and $x_{ki}$ are respectively, given by

$$x_{kr}=\text{Re}\left(x_k\right)=\cos \left(\frac{(2k-1)\pi}{n}\right) \tag 2$$

$$x_{ki}=\text{Im}\left(x_k\right)=\sin \left(\frac{(2k-1)\pi}{n}\right) \tag 3$$

Using $(1)$-$(3)$ with $n=3$, we find that

$$\int \frac{1}{1+x^3}\,dx=-\frac16 \left(\log(x^2-x+1)-\log(x^2+2x+1)\right)+\frac{\sqrt{3}}{3}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+C$$

Simplifying, we obtain

$$\int \frac{1}{1+x^3}\,dx=-\frac16 \log(x^2-x+1)+\frac13\log(x+1)+\frac{\sqrt{3}}{3}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+C$$

Answer 4

You can factor the denominator using the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$, and in our case $a=x$ and $b=1$

$$\int \dfrac{1}{(x+1)(x^2-x+1)} dx$$

Assume $\dfrac{1}{(x+1)(x^2-x+1)} = \dfrac {A}{x+1} + \dfrac {Bx+C}{x^2-x+1} $

Can you take it from here?

Answer 5

Here is a roundabout way. Let $$I:=\int\frac1{1+x^3}dx, J:=\int\frac x{1+x^3}dx,$$ then

$$I+J=\int\frac {1+x}{1+x^3}dx=\int\frac 1{1-x+x^2}dx=\int\frac1{\left(x-\frac1{2}\right)^2+\frac{3}{4}}dx=\frac2{\sqrt3}\arctan\frac{2x-1}{\sqrt3}+C,$$

$$I-J=\int\frac{1-x}{1+x^3}dx=\int\frac1{1+x}dx-\int\frac{x^2}{1+x^3}dx=\ln\lvert1+x\rvert-\frac1{3}\ln\lvert 1+x^3\rvert+C,$$ solving the above equations yields $$I=\frac1{\sqrt{3}}\arctan\frac{2x-1}{\sqrt3}+\frac1{2}\ln\lvert1+x\rvert-\frac1{6}\ln\lvert1+x^3\rvert+C,$$ $$J=\frac1{\sqrt{3}}\arctan\frac{2x-1}{\sqrt3}-\frac1{2}\ln\lvert1+x\rvert+\frac1{6}\ln\lvert1+x^3\rvert+C.$$