When solving integrals with rational functions, substitution is often useful, especially when the denominator involves a quadratic expression. Pay close attention to how the limits of integration change when performing the substitution, as this ensures the integral remains properly evaluated. For terms like \( \frac{1}{u^2} \), remember that integrating powers of \( u \) involves basic power rule applications, and logarithmic integrals like \( \frac{1}{u} \) lead to natural logarithms. Always simplify the expression step by step for better clarity in your solution.
The initial integral is:
\[ \int_{0}^{1} \frac{a - bx^2}{(a + bx^2)^2} dx \]
Let \( u = a + bx^2 \). Then \( du = 2bx \, dx \), which implies \( x \, dx = \frac{du}{2b} \).
The integration limits are transformed as follows:
When \( x = 0 \), \( u = a \). When \( x = 1 \), \( u = a + b \).
Substituting these into the integral yields:
\[ \int_{a}^{a+b} \frac{2a - u}{u^2} \cdot \frac{1}{2b} \, du. \]
This simplifies to:
\[ \frac{1}{2b} \int_{a}^{a+b} \left( \frac{2a}{u^2} - \frac{1}{u} \right) du. \]
Evaluating the first term, \( \int_{a}^{a+b} \frac{2a}{u^2} du \):
\[ \int \frac{2a}{u^2} du = -\frac{2a}{u}. \]
The evaluation of this term is:
\[ -\frac{2a}{a + b} + \frac{2a}{a}. \]
Evaluating the second term, \( \int_{a}^{a+b} \frac{1}{u} du \):
\[ \ln(a + b) - \ln(a). \]
Combining the results:
\[ \frac{1}{2b} \left( -\frac{2a}{a + b} + 2 - (\ln(a + b) - \ln(a)) \right). \]
The final simplified result is:
\[ \int_{0}^{1} \frac{a - bx^2}{(a + bx^2)^2} dx = \frac{1}{a + b}. \]