Question

Evaluate the integral $\int_0^1 \frac{a - bx^2}{(a + bx^2)^2} , dx$:

• A. $\frac{a - b}{a + b}$
• B. $\frac{1}{a - b}$
• C. $\frac{a + b}{2}$
• D. $\frac{1}{a + b}$

Answer 1

The Correct Option is D

We start with the integral:

\[ \int_{0}^{1} \frac{a - bx^2}{(a + bx^2)^2} dx \]

Let \( u = a + bx^2 \), so:

\[ du = 2bx \, dx \quad \text{and} \quad x \, dx = \frac{du}{2b}. \]

The limits change as follows:

For \( x = 0 \Rightarrow u = a \) and for \( x = 1 \Rightarrow u = a + b \).

Substitute into the integral:

\[ \int_{a}^{a+b} \frac{2a - u}{u^2} \cdot \frac{1}{2b} \, du. \]

Simplify:

\[ \frac{1}{2b} \int_{a}^{a+b} \left( \frac{2a}{u^2} - \frac{1}{u} \right) du. \]

Solve each term: For \( \int_{a}^{a+b} \frac{2a}{u^2} du \):

\[ \int \frac{2a}{u^2} du = -\frac{2a}{u}. \]

Evaluate:

\[ -\frac{2a}{a + b} + \frac{2a}{a}. \]

For \( \int_{a}^{a+b} \frac{1}{u} du \):

\[ \ln(a + b) - \ln(a). \]

Combine results:

\[ \frac{1}{2b} \left( -\frac{2a}{a + b} + 2 - (\ln(a + b) - \ln(a)) \right). \]

Final simplification gives:

\[ \int_{0}^{1} \frac{a - bx^2}{(a + bx^2)^2} dx = \frac{1}{a + b}. \]