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

##### 1 Answer

# 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

# 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

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#