How do you evaluate the integral #int 1/(xsqrt(1-x^2))#?

Question

19607 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-30T07:03:55

The answer is# =-1/2ln(sqrt(1-x^2)+1)+1/2ln(|sqrt(1-x^2)-1|)+C#

We perform this integral by substitution

Let #u=sqrt(1-x^2)#

#u^2=1-x^2#

#du=(-2xdx)/(2sqrt(1-x^2))=(-xdx)/(sqrt(1-x^2))#

Therefore,

#intdx/(xsqrt(1-x^2))=intsqrt(1-x^2)(du)/(-x^2sqrt(1-x^2))#

#=int-(du)/x^2#

#=int(du)/(u^2-1)#

Now, we perform a de composition 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))#

The denominators are the same, we can compare the numerators

#1=A(u-1)+B(u+1)#

Let #u=-1#, #=>#, #1=-2A#, #=>#, #A=-1/2#

Let #u=1#, #=>#, #1=2B#, #=>#, #B=1/2#

Therefore,

#1/(u^2-1)=(-1/2)/(u+1)+(1/2)/(u-1)#

So,

#int(du)/(u^2-1)=-1/2int(du)/(u+1)+1/2int(du)/(u-1)#

#=-1/2ln(u+1)+1/2ln(u-1)#

#=-1/2ln(sqrt(1-x^2)+1)+1/2ln(|sqrt(1-x^2)-1|)+C#