0

Evaluation of $\displaystyle \int\frac{1}{1+x^6}dx$

$\bf{My\; Try::}$Let $$I = \int\frac{1}{1+x^6}dx = \int\frac{1}{(1+x^2)(x^4-x^2+1)}dx$$

Using Partial fraction , above Integral is very lengthy, Can we solve it without using

partial fraction or any other way, Help me

Thanks

Martin Sleziak
  • 53,687
juantheron
  • 53,015

2 Answers2

6

I would also recommend partial fractions but instead with

$$\dfrac{Ax+b}{x^2+\sqrt{3}x+1}+\dfrac{Cx+D}{x^2-\sqrt{3}x+1}+\dfrac{Ex+F}{x^2+1}$$

Claire
  • 165
2

$$\displaystyle \int\frac{1}{1+x^6}dx = \frac{1}{2}\int\frac{\left(1+x^4\right)+\left(1-x^4\right)}{1+x^6}dx$$

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

Now we will take $\displaystyle I = \frac{1}{2}\int\frac{1+x^4}{1+x^6}dx$ and $\displaystyle J = \frac{1}{2}\int\frac{1-x^4}{1+x^6}dx$

So first we will calculate value of $I$

So $$\displaystyle I = \frac{1}{2}\int\frac{1+x^4}{1+x^6}dx = \frac{1}{2}\int\frac{(x^2+1)^2-2x^2}{1+x^6}dx$$

So $$\displaystyle I = \frac{1}{2}\int\frac{(x^2+1)^2}{(1+x^2)\cdot (x^4-x^2+1)}dx - \int\frac{x^2}{1+(x^3)^2}dx$$

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

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

Now Let $\displaystyle \left(x-\frac{1}{x}\right) = t \Leftrightarrow \left(1+\frac{1}{x^2}\right)dx = dt$ and $x^3 = u\Leftrightarrow 3x^2dx = du\displaystyle \Leftrightarrow dx = \frac{1}{3}du$

So $$\displaystyle I = \frac{1}{2}\cdot \tan^{-1}\left(x-\frac{1}{x}\right) - \frac{1}{3}\cdot \tan^{-1}\left(x^3\right)+\mathcal{C'}$$

Similarly we will calculate for $$\displaystyle J = \frac{1}{2}\int\frac{1-x^4}{1+x^6}dx$$

So $$\displaystyle J = \frac{1}{2}\int\frac{(1-x^2)\cdot (1+x^2)}{(1+x^2)\cdot (x^4-x^2+1)}dx = -\frac{1}{2}\int\frac{x^2-1}{x^4-x^2+1}dx$$

$$\displaystyle J = -\frac{1}{2}\int \frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}\right)^2+\left(\sqrt{3}\right)^2}dx$$

Now Now Let $\displaystyle \left(x+\frac{1}{x}\right) = v \Leftrightarrow \left(1-\frac{1}{x^2}\right)dx = dv$

So $$\displaystyle J = -\frac{1}{2}\cdot \frac{1}{2\sqrt 3}\cdot \ln \left|\frac{\left(x+\frac{1}{x}\right)-\sqrt{3}}{\left(x+\frac{1}{x}\right)+\sqrt{3}}\right|+\mathcal{C''}$$

So $\displaystyle \int \frac{1}{1+x^6}dx = \frac{1}{2}\cdot \tan^{-1}\left(x-\frac{1}{x}\right) - \frac{1}{3}\cdot \tan^{-1}\left(x^3\right) - \frac{1}{4\sqrt{3}}\cdot \ln \left|\frac{\left(x+\frac{1}{x}\right)-\sqrt{3}}{\left(x+\frac{1}{x}\right)+\sqrt{3}}\right|+\mathcal{C}$

juantheron
  • 53,015