Question

The integral $\int^{^{\frac{1}{2}}}_{_0} \frac{ln \left(1+2x\right)}{1+4x^{2}}dx,$ equals :

• A. $\frac{\pi}{4}$ ln 2
• B. $\frac{\pi}{8}$ ln 2
• C. $\frac{\pi}{16}$ ln 2
• D. $\frac{\pi}{32}$ ln 2

Answer 1

The Correct Option is C

To solve the integral $\int^{\frac{1}{2}}_{0} \frac{\ln(1 + 2x)}{1 + 4x^2} \, dx$, we will use the properties of definite integrals and substitutions to simplify the expression.

Step 1: Recognizing Symmetry

The integral is of the form $\int_0^a f(x) \, dx$ where some symmetry can be identified. Recognize the integral $\int^{\frac{1}{2}}_{0} \frac{\ln(1 + 2x)}{1 + 4x^2} \, dx$ and try to find a substitution that simplifies it.

Step 2: Substitution and Simplification

Perform the substitution $x = \frac{1}{2} \tan(\theta)$, which implies $dx = \frac{1}{2} \sec^2(\theta) \, d\theta$. The limits of integration change to:

• When $x = 0$, $\theta = 0$
• When $x = \frac{1}{2}$, $\theta = \frac{\pi}{4}$

The integral becomes:

$$\int_0^{\frac{\pi}{4}} \frac{\ln(1 + \tan(\theta))}{\sec^2 (\theta)} \cdot \frac{1}{2} \sec^2(\theta) \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{4}} \ln(1 + \tan(\theta)) \, d\theta$$

Step 3: Solving the Integral

The function $\ln(1 + \tan(\theta))$ can be approached using symmetry or known results. Consider the particular symmetry of definite integrals and previous solved results:

We know that:

$$ \int_0^{\frac{\pi}{4}} \ln(1 + \tan(\theta)) \, d\theta = \frac{\pi}{8} \ln 2 $$

Now multiplying the result by $\frac{1}{2}$, we find:

$$ \frac{1}{2} \int_0^{\frac{\pi}{4}} \ln(1 + \tan(\theta)) \, d\theta = \frac{1}{2} \cdot \frac{\pi}{8} \ln 2 = \frac{\pi}{16} \ln 2 $$

Conclusion

Hence, the integral $\int^{\frac{1}{2}}_{0} \frac{\ln(1 + 2x)}{1 + 4x^2} \, dx$ evaluates to $\frac{\pi}{16}$ ln 2. Therefore, the correct option is $\frac{\pi}{16}$ ln 2.