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

Answer: (D)

$\frac{1}{a + b}$

Answer 2

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:

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