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

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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,

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

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

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

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