Question

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

Answer 1

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

Explanation:

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`