# int 1/(1+x^3) dx is equal to?

May 2, 2018

$- \frac{1}{6} \left(\ln | {x}^{2} - x - 1 | - 2 \sqrt{3} \arctan \left(\frac{2 x - 1}{\sqrt{3}}\right) - 2 \ln | x + 1 |\right) + C$

#### Explanation:

If we say $F ' \left(x\right) = \frac{1}{1 + {x}^{3}}$,
we want to find F(x) where $\int F ' \left(x\right) \mathrm{dx} = F \left(x\right) + C$.
To simplify I am skipping the constant C.

I will suppose you have a list of standard integration expressions, like in https://en.wikipedia.org/wiki/Lists_of_integrals
Therefore, I will suppose you know that
$\int \frac{1}{x} \mathrm{dx} = \ln | x |$
and
$\int \frac{1}{{x}^{2} + 1} \mathrm{dx} = \arctan x$
I will, therefore, take these as given.

$\int \frac{1}{{x}^{3} + 1} \mathrm{dx}$ (plase the unknown first)
=$\int \frac{1}{\left(x + 1\right) \left({x}^{2} - x + 1\right)} \mathrm{dx}$ (factorise the denominator)

We observe that $\left(x + 1\right) \left(x - 2\right) = \left({x}^{2} - x - 2\right) = \left({x}^{2} - x + 1\right) - 3$
So $\left({x}^{2} - x + 1\right) - \left(x + 1\right) \left(x - 2\right) = 3$

We, therefore, can write our expression as:
$\int \frac{3}{3 \left(x + 1\right) \left({x}^{2} - x + 1\right)} \mathrm{dx} = \int \frac{\left({x}^{2} - x + 1\right) - \left(x + 1\right) \left(x - 2\right)}{3 \left(x + 1\right) \left({x}^{2} - x + 1\right)}$ which we can split in two:
(1) $= \frac{1}{3} \int \frac{1}{x + 1} \mathrm{dx} - \frac{1}{3} \int \frac{x - 2}{{x}^{2} - x + 1} \mathrm{dx}$

In the first part in (1) we see:
$\int \frac{1}{x + 1} \mathrm{dx} = \int \frac{1}{u} \mathrm{du}$ (where u=x+1)
(2) =$\ln \left(u\right) = \ln \left(x + 1\right)$

In the second part in (1) we have:
If $u = {x}^{2} - x + 1$
we get $u ' = 2 x - 1$
i.e. $\mathrm{dx} = \frac{1}{2 x - 1} \mathrm{du}$
In addition we have $x - 2 = \frac{1}{2} \left(2 x - 1\right) - \frac{3}{2}$

Using this we get in (1):
$\int \frac{x - 2}{{x}^{2} - x + 1} \mathrm{dx}$
(3) =$\frac{1}{2} \int \frac{2 x - 1}{{x}^{2} - x + 1} \mathrm{dx} - \frac{3}{2} \int \left(\frac{1}{{x}^{2} - x + 1}\right) \mathrm{dx}$

Substitute $u = {x}^{2} - x + 1$, $\mathrm{dx} = \frac{1}{2 x - 1} \mathrm{du}$ in the first part:
$\int \frac{2 x - 1}{{x}^{2} - x + 1} \mathrm{dx}$
(4) =$\int \frac{1}{u} \mathrm{du} = \ln \left(u\right) = \ln \left({x}^{2} - x + 1\right)$

The second part in (3):
$\int \frac{1}{{x}^{2} - x + 1} \mathrm{dx}$

The numerator can be written:
${\left(x - \frac{1}{2}\right)}^{2} + \frac{3}{4}$
Therefore
$\int \frac{1}{{x}^{2} - x + 1} \mathrm{dx} = \int \frac{1}{{\left(x - \frac{1}{2}\right)}^{2} + \frac{3}{4}} \mathrm{dx}$
If we substitue $u = \frac{2 x - 1}{\sqrt{3}}$, $\mathrm{dx} = \frac{\sqrt{3}}{2} \mathrm{du}$ in this, we get
$= \int \frac{2 \sqrt{3}}{3} {u}^{2} + 3 = \frac{2}{\sqrt{3}} \int \frac{1}{{u}^{2} + 1} \mathrm{du}$
$= \frac{2}{\sqrt{3}} \arctan \left(u\right)$
$= \frac{2}{\sqrt{3}} \arctan \left(\frac{2 x - 1}{\sqrt{3}}\right)$

Plug this into (3):
(3) =$\frac{1}{2} \int \frac{2 x - 1}{{x}^{2} - x + 1} \mathrm{dx} - \frac{3}{2} \int \left(\frac{1}{{x}^{2} - x + 1}\right) \mathrm{dx}$
$= \frac{1}{2} \ln \left({x}^{2} - x + 1\right) - \sqrt{3} \arctan \left(\frac{2 x - 1}{\sqrt{3}}\right)$

Plug this into (1):
$= \frac{1}{3} \int \frac{1}{x + 1} \mathrm{dx} - \frac{1}{3} \int \frac{x - 2}{{x}^{2} - x + 1} \mathrm{dx}$
$= - \ln \frac{{x}^{2} - x + 1}{6} + \arctan \left(\frac{2 x - 1}{\sqrt{3}}\right) + \ln \frac{x + 1}{3}$

Last step: apply the absolute value to the arguments of the logarithm functions:
$= - \frac{\ln | {x}^{2} - x + 1 |}{6} + \arctan \left(\frac{2 x - 1}{\sqrt{3}}\right) + \ln | x + 1 \frac{|}{3}$

May 2, 2018

$I = \frac{1}{3} \ln | x + 1 | - \frac{1}{6} \ln | {x}^{2} - x + 1 | + \frac{1}{\sqrt{3}} {\tan}^{-} 1 \left(\frac{2 x - 1}{\sqrt{3}}\right) + c$

#### Explanation:

Here,

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

$= \int \frac{1}{\left(x + 1\right) \left({x}^{2} - x + 1\right)} \mathrm{dx}$

We have,

$\frac{1}{\left(x + 1\right) \left({x}^{2} - x + 1\right)} = \frac{A}{x + 1} + \frac{B x + C}{{x}^{2} - x + 1}$

1=A(x^2-x+1)+(Bx+C)((x+1)

$x = - 1 \implies 1 = A \left(1 + 1 + 1\right) \implies A = \frac{1}{3}$

$x = 0 \implies 1 = A + C \implies C = 1 - A = 1 - \frac{1}{3} \implies C = \frac{2}{3}$

Also,

$1 = {x}^{2} \left(A + B\right) + x \left(- A + B + C\right) + \left(A + C\right)$

Comparing coefficient of ${x}^{2}$,both sides

$A + B = 0 \implies \frac{1}{3} + B = 0 \implies B = - \frac{1}{3}$

So,

$I = \int \frac{\frac{1}{3}}{x + 1} \mathrm{dx} + \int \left(\frac{- \frac{1}{3} x + \frac{2}{3}}{{x}^{2} - x + 1}\right) \mathrm{dx}$

$= \frac{1}{3} \int \frac{1}{x + 1} \mathrm{dx} - \frac{1}{3} \int \frac{x - 2}{{x}^{2} - x + 1} \mathrm{dx}$

$= \frac{1}{3} \int \frac{1}{x + 1} \mathrm{dx} - \frac{1}{6} \int \frac{2 x - 4}{{x}^{2} - x + 1} \mathrm{dx}$

=1/3int1/(x+1)dx-1/6int(2x-1)/(x^2-x+1)dx-1/6int(-3)/(x^2- x+1)dx

=1/3ln|x+1|-1/6int(d/(dx)(x^2-x+1))/(x^2-x+1)dx+1/2int1/((x- 1/2)^2+(sqrt3/2)^2)dx

=1/3ln|x+1|-1/6ln|x^2-x+1|+1/2(2/sqrt3)tan^-1((x-1/2)/(sqrt3/2)

$= \frac{1}{3} \ln | x + 1 | - \frac{1}{6} \ln | {x}^{2} - x + 1 | + \frac{1}{\sqrt{3}} {\tan}^{-} 1 \left(\frac{2 x - 1}{\sqrt{3}}\right) + c$

Note:
$A , B , \mathmr{and} C$ can also be determine by comparing coefficient of ${x}^{2} , x \mathmr{and}$ constant terms of both sides.