Evaluate $ \int_{ \minus{} 1}^ {a^2} \frac {1}{x^2 \plus{} a^2}\ dx\ (a > 0).$

You may not use $ \tan ^{ \minus{} 1} x$ or Complex Integral here.
To evaluate the integral \( \int_{-1}^{a^2} \frac{1}{x^2 + a^2} \, dx \) without using \( \tan^{-1} x \) or complex integrals, we can use a substitution method.

1. **Substitution**:
   Let \( x = a \tan(\theta) \). Then, \( dx = a \sec^2(\theta) \, d\theta \).

2. **Change of Limits**:
   When \( x = -1 \):
   \[
   -1 = a \tan(\theta) \implies \tan(\theta) = -\frac{1}{a} \implies \theta = \tan^{-1}\left(-\frac{1}{a}\right)
   \]
   When \( x = a^2 \):
   \[
   a^2 = a \tan(\theta) \implies \tan(\theta) = a \implies \theta = \tan^{-1}(a)
   \]

3. **Transform the Integral**:
   Substitute \( x = a \tan(\theta) \) into the integral:
   \[
   \int_{-1}^{a^2} \frac{1}{x^2 + a^2} \, dx = \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} \frac{1}{a^2 \tan^2(\theta) + a^2} \cdot a \sec^2(\theta) \, d\theta
   \]
   Simplify the integrand:
   \[
   \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} \frac{a \sec^2(\theta)}{a^2 (\tan^2(\theta) + 1)} \, d\theta = \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} \frac{a \sec^2(\theta)}{a^2 \sec^2(\theta)} \, d\theta = \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} \frac{1}{a} \, d\theta
   \]
   \[
   = \frac{1}{a} \int_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} d\theta
   \]

4. **Evaluate the Integral**:
   \[
   \frac{1}{a} \left[ \theta \right]_{\tan^{-1}(-\frac{1}{a})}^{\tan^{-1}(a)} = \frac{1}{a} \left( \tan^{-1}(a) - \tan^{-1}(-\frac{1}{a}) \right)
   \]

5. **Simplify the Result**:
   Recall that \( \tan^{-1}(-x) = -\tan^{-1}(x) \):
   \[
   \frac{1}{a} \left( \tan^{-1}(a) + \tan^{-1}(\frac{1}{a}) \right)
   \]
   Using the identity \( \tan^{-1}(x) + \tan^{-1}(\frac{1}{x}) = \frac{\pi}{2} \) for \( x > 0 \):
   \[
   \frac{1}{a} \cdot \frac{\pi}{2} = \frac{\pi}{2a}
   \]

The final answer is \(\boxed{\frac{\pi}{2a}}\).