Evaluate the integral? : $\int \frac{1}{x \sqrt{1 - x^{4}}} d x$ $int 1/(xsqrt(1-x^4)) dx$

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Answer 1 · 2017-10-03T21:16:44

$int \ 1/(xsqrt(1-x^4)) \ dx = 1/4 (ln|sqrt(1-x^4)-1| -ln|sqrt(1-x^4)+1 )+ C$

We seek:

$I = 1/(xsqrt(1-x^4)) \ dx$

Let us attempt an substitution of the form:

$u = sqrt(1-x^4) => u^2 = 1-x^4$ , and $x^4 = 1-u^2$

And differentiating wrt $x$ we get:

$2u(du)/dx = -4x^3 => (2u)/(-4x^4) (du)/dx = (-4x^3)/(-4x^4)$
$:. -(u)/(2x^4) (du)/dx = 1/x$

If we substitute this into the integral, we get:

$I = int \ (1/u) ( -(u)/(2(1-u^2))) \ du$
$\ \ = 1/2 \ int \ 1/(u^2-1) \ du$

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

$1/(u^2-1) -= 1/((u+1)(u-1))$
$\ \ \ \ \ \ \ \ \ \ \ \ = A/(u+1) + B/(u-1)$
$\ \ \ \ \ \ \ \ \ \ \ \ = ( A(u-1) + B(u+1) ) / ((u+1)(u-1))$

Leadin to:

$1 = (u-1) + B(u+1)$

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 $u = +1 => 0 = 2B => B=1/2$
Put $u = -1 => 0 = -2A => A=-1/2$

Thus:

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

Which we can directly integrate:

$I = 1/4 (ln|u-1| -0 ln|u+1 )+ C$

Restoring the substitution:

$I = 1/4 (ln|sqrt(1-x^4)-1| -ln|sqrt(1-x^4)+1 )+ C$

Evaluate the integral? : int 1/(xsqrt(1-x^4)) dx∫1x√1−x4dx int 1/(xsqrt(1-x^4)) dx