Answer: (a) \(\frac{dy}{dx}=\frac{2x}{1-x^2}\), so
\(1+\left(\frac{dy}{dx}\right)^2=1+\frac{4x^2}{(1-x^2)^2}=\frac{(1-x^2)^2+4x^2}{(1-x^2)^2}=\frac{1+2x^2+x^4}{(1-x^2)^2}=\left(\frac{1+x^2}{1-x^2}\right)^2.\)
(b) We seek constants \(P,Q,R\) such that
\(\frac{1+x^2}{1-x^2}=\frac{P}{1+x}+\frac{Q}{1-x}+R.\)
Since \(1-x^2=(1-x)(1+x)\), write
\(1+x^2=P(1-x)+Q(1+x)+R(1-x^2).\)
Expanding and comparing coefficients gives \(P=1, Q=1, R=-1\). Hence
\(\frac{1+x^2}{1-x^2}=\frac{1}{1+x}+\frac{1}{1-x}-1.\)
The arc length is
\(s=\int_{-1/2}^{1/2}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=\int_{-1/2}^{1/2}\frac{1+x^2}{1-x^2}\,dx.\)
Using the symmetry of the integrand,
\(s=2\int_0^{1/2}\left(\frac{1}{1+x}+\frac{1}{1-x}-1\right)dx.\)
Now
\(\int\left(\frac{1}{1+x}+\frac{1}{1-x}-1\right)dx=\ln(1+x)-\ln(1-x)-x=\ln\left(\frac{1+x}{1-x}\right)-x.\)
So
\(s=2\left[\ln\left(\frac{1+x}{1-x}\right)-x\right]_0^{1/2}=2\left(\ln 3-\frac12\right)=2\ln 3-1.\)
Therefore the exact arc length is \(2\ln 3-1\).
(a) Differentiate \(y=-\ln(1-x^2)\):
\(\frac{dy}{dx}=-\frac{1}{1-x^2}\cdot(-2x)=\frac{2x}{1-x^2}.\)
Then
\(1+\left(\frac{dy}{dx}\right)^2=1+\frac{4x^2}{(1-x^2)^2}.\)
Put over a common denominator:
\(1+\left(\frac{dy}{dx}\right)^2=\frac{(1-x^2)^2+4x^2}{(1-x^2)^2}=\frac{1-2x^2+x^4+4x^2}{(1-x^2)^2}.\)
Simplifying the numerator gives
\(1+2x^2+x^4=(1+x^2)^2,\)
so
\(1+\left(\frac{dy}{dx}\right)^2=\left(\frac{1+x^2}{1-x^2}\right)^2.\)
Hence, since the integrand for arc length is positive on \([-frac12,frac12]\),
\(\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\frac{1+x^2}{1-x^2}.\)
(b) Now express \(\frac{1+x^2}{1-x^2}\) in partial fractions. Write
\(\frac{1+x^2}{1-x^2}=\frac{P}{1+x}+\frac{Q}{1-x}+R.\)
Multiplying through by \(1-x^2=(1-x)(1+x)\) gives
\(1+x^2=P(1-x)+Q(1+x)+R(1-x^2).\)
Expand:
\(1+x^2=(P+Q+R)+( -P+Q)x-Rx^2.\)
Comparing coefficients with \(1+0x+x^2\):
- constant term: \(P+Q+R=1\)
- \(x\)-term: \(-P+Q=0\)
- \(x^2\)-term: \(-R=1\)
Thus \(R=-1\), \(Q=P\), and \(2P-1=1\), so \(P=1\) and \(Q=1\). Therefore
\(\frac{1+x^2}{1-x^2}=\frac{1}{1+x}+\frac{1}{1-x}-1.\)
For the arc length \(s\), use
\(s=\int_{-1/2}^{1/2}\frac{1+x^2}{1-x^2}\,dx.\)
The integrand is even, so
\(s=2\int_0^{1/2}\left(\frac{1}{1+x}+\frac{1}{1-x}-1\right)dx.\)
Integrate term by term:
\(\int \frac{1}{1+x}\,dx=\ln(1+x),\)
\(\int \frac{1}{1-x}\,dx=-\ln(1-x),\)
and \(\int 1\,dx=x\). Hence
\(s=2\left[\ln(1+x)-\ln(1-x)-x\right]_0^{1/2}=2\left[\ln\left(\frac{1+x}{1-x}\right)-x\right]_0^{1/2}.\)
Substitute \(x=frac12\):
\(\ln\left(\frac{1+1/2}{1-1/2}\right)=\ln 3,\)
so
\(s=2\left(\ln 3-\frac12\right)=2\ln 3-1.\)
Therefore the exact arc length is \(2\ln 3-1\).
