# Evaluate the integral? :  int 1/(xsqrt(1-x^4)) dx

Oct 3, 2017

$\int \setminus \frac{1}{x \sqrt{1 - {x}^{4}}} \setminus \mathrm{dx} = \frac{1}{4} \left(\ln | \sqrt{1 - {x}^{4}} - 1 | - \ln | \sqrt{1 - {x}^{4}} + 1\right) + C$

#### Explanation:

We seek:

$I = \frac{1}{x \sqrt{1 - {x}^{4}}} \setminus \mathrm{dx}$

Let us attempt an substitution of the form:

$u = \sqrt{1 - {x}^{4}} \implies {u}^{2} = 1 - {x}^{4}$, and ${x}^{4} = 1 - {u}^{2}$

And differentiating wrt $x$ we get:

$2 u \frac{\mathrm{du}}{\mathrm{dx}} = - 4 {x}^{3} \implies \frac{2 u}{- 4 {x}^{4}} \frac{\mathrm{du}}{\mathrm{dx}} = \frac{- 4 {x}^{3}}{- 4 {x}^{4}}$
$\therefore - \frac{u}{2 {x}^{4}} \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

If we substitute this into the integral, we get:

$I = \int \setminus \left(\frac{1}{u}\right) \left(- \frac{u}{2 \left(1 - {u}^{2}\right)}\right) \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{1}{{u}^{2} - 1} \setminus \mathrm{du}$

Now, we can decompose this new integrand into partial fractions:

$\frac{1}{{u}^{2} - 1} \equiv \frac{1}{\left(u + 1\right) \left(u - 1\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{A}{u + 1} + \frac{B}{u - 1}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{A \left(u - 1\right) + B \left(u + 1\right)}{\left(u + 1\right) \left(u - 1\right)}$

$1 = \left(u - 1\right) + B \left(u + 1\right)$

Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $u = + 1 \implies 0 = 2 B \implies B = \frac{1}{2}$
Put $u = - 1 \implies 0 = - 2 A \implies A = - \frac{1}{2}$

Thus:

$I = \frac{1}{2} \setminus \int \setminus \frac{1}{{u}^{2} - 1} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{2} \setminus \int \setminus \left\{\frac{- \frac{1}{2}}{u + 1} + \frac{\frac{1}{2}}{u - 1}\right\} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{4} \setminus \int \setminus \left\{\frac{1}{u - 1} - \frac{1}{u + 1}\right\} \setminus \mathrm{du}$

Which we can directly integrate:

$I = \frac{1}{4} \left(\ln | u - 1 | - 0 \ln | u + 1\right) + C$

Restoring the substitution:

$I = \frac{1}{4} \left(\ln | \sqrt{1 - {x}^{4}} - 1 | - \ln | \sqrt{1 - {x}^{4}} + 1\right) + C$