Question

The value of the integral $\int_{\ln 2}^{\ln 3} \frac{e^{2x} - 1}{e^{2x} + 1} \, dx$ is:

• A. $\ln 3$
• B. $\ln 4 - \ln 3$
• C. $\ln 9 - \ln 4$
• D. $\ln 3 - \ln 2$

Answer 1

Answer: (B)

$\ln 4 - \ln 3$

Answer 2

Let:

I = $\int_{\log_{e} 2}^{\log_{e} 3} \frac{e^{2x} - 1}{e^{2x} + 1} dx$.

Substitute $u = e^{2x}$, so $du = 2e^{2x}dx$ or $dx = \frac{du}{2u}$. The limits become:

$x = \log_{e} 2 \implies u = e^{2 \cdot \log_{e} 2} = 4$,

$x = \log_{e} 3 \implies u = e^{2 \cdot \log_{e} 3} = 9$.

The integral becomes:

I = $\frac{1}{2} \int_{4}^{9} \frac{u - 1}{u(u + 1)} du$.

Split the fraction:

$\frac{u - 1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1}$.

Thus:

I = $\frac{1}{2} \int_{4}^{9} \left( \frac{1}{u} - \frac{1}{u + 1} \right) du$.

Integrate:

I = $\frac{1}{2} [\ln u - \ln(u + 1)]_{4}^{9}$.

Simplify:

I = $\frac{1}{2} \left[(\ln 9 - \ln 10) - (\ln 4 - \ln 5)\right]$.

Combine terms:

I = $\frac{1}{2} \left[\ln \left(\frac{9}{10}\right) - \ln \left(\frac{4}{5}\right)\right]$.

Simplify further:

I = $\frac{1}{2} \ln \left(\frac{9}{10} \times \frac{5}{4}\right)$.

Simplify:

I = $\frac{1}{2} \ln \left(\frac{45}{40}\right)$.

This simplifies to:

I = $\ln 4 - \ln 3$.

Thus:

$\ln_{e} 4 - \ln_{e} 3$.