Calculate the integral: integral from - 1/2 to (1/2)

Calculate the integral: $\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{dx}{1-x^2}$.

We will use the Newton-Leibniz formula after finding the antiderivative using partial fractions.

Step 1: Factor the denominator and use partial fractions

\[\frac{1}{1-x^2} = \frac{1}{(1-x)(1+x)} = \frac{A}{1-x} + \frac{B}{1+x}\]

Multiplying both sides by $(1-x)(1+x)$: $1 = A(1+x) + B(1-x)$.

Setting $x=1$: $1=2A \Rightarrow A=\frac12$. Setting $x=-1$: $1=2B \Rightarrow B=\frac12$.

Therefore:

\[\frac{1}{1-x^2} = \frac{1}{2} \cdot \frac{1}{1-x} + \frac{1}{2} \cdot \frac{1}{1+x}\]

Step 2: Find the antiderivative

\[\int \frac{1}{1-x^2} \, dx = \frac{1}{2} \int \frac{1}{1-x} \, dx + \frac{1}{2} \int \frac{1}{1+x} \, dx\]

For the first integral, let $u=1-x$, so $du=-dx$: $\int \frac{1}{1-x}\,dx = -\ln|1-x|$.

For the second integral: $\int \frac{1}{1+x}\,dx = \ln|1+x|$.

Therefore:

\[\int \frac{1}{1-x^2} \, dx = \frac{1}{2}\ln|1+x| - \frac{1}{2}\ln|1-x| + C = \frac{1}{2}\ln\left|\frac{1+x}{1-x}\right| + C\]

Step 3: Apply the Newton-Leibniz formula

\[\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{dx}{1-x^2} = \left[\frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|\right]_{-\frac{1}{2}}^{\frac{1}{2}}\]

Step 4: Evaluate at the upper limit $x = \frac{1}{2}$

\[F\left(\frac{1}{2}\right) = \frac{1}{2}\ln\left|\frac{3/2}{1/2}\right| = \frac{1}{2}\ln(3)\]

Step 5: Evaluate at the lower limit $x = -\frac{1}{2}$

\[F\left(-\frac{1}{2}\right) = \frac{1}{2}\ln\left|\frac{1/2}{3/2}\right| = \frac{1}{2}\ln\left(\frac{1}{3}\right) = -\frac{1}{2}\ln(3)\]

Step 6: Calculate the definite integral

\[\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{dx}{1-x^2} = F\left(\frac{1}{2}\right) - F\left(-\frac{1}{2}\right) = \frac{1}{2}\ln(3) - \left(-\frac{1}{2}\ln(3)\right) = \ln(3)\]

Alternative approach using symmetry:

Since $f(x)=\frac{1}{1-x^2}$ is even:

\[\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{dx}{1-x^2} = 2\int_0^{\frac{1}{2}} \frac{dx}{1-x^2} = \left[\ln\left|\frac{1+x}{1-x}\right|\right]_0^{\frac{1}{2}} = \ln(3) - \ln(1) = \ln(3)\]

This confirms our result!

Geometric interpretation:

The function $\frac{1}{1-x^2}$ has vertical asymptotes at $x=\pm1$, is positive for $|x|<1$, and is symmetric about the $y$-axis. At $x=0$: $y=1$ (minimum in the interval); at $x=\pm\frac12$: $y=\frac{4}{3}$. The function increases as we move away from $x=0$ toward the asymptotes.

Numerical value:

Using $\ln(3) \approx 1.0986$: $\int_{-1/2}^{1/2} \frac{dx}{1-x^2} \approx 1.0986$.

Final answer:

\[\boxed{\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{dx}{1-x^2} = \ln(3)}\]