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

Hint:

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.

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:

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

Answer 2

We start with the integral:

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

Step 1: Use substitution $u = a + bx^2$:

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

Step 2: Change the limits of integration:

For $x = 0 \Rightarrow u = a$, and for $x = 1 \Rightarrow u = a + b$.

Step 3: Substitute into the integral:

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

Step 4: Simplify the expression:

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

Step 5: Solve each term:

For $\int_{a}^{a+b} \frac{2a}{u^2} du$, we know: $$\int \frac{2a}{u^2} du = -\frac{2a}{u}.$$ Evaluating: $$-\frac{2a}{a + b} + \frac{2a}{a}.$$ For $\int_{a}^{a+b} \frac{1}{u} du$, we get: $$\ln(a + b) - \ln(a).$$

Step 6: Combine results:

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

Step 7: Final simplification:

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