CUET Mathematics 2024 - The value of the integral _ _e 2^ _e 3 e^2 x-1e^2 x+1 d x is : | PYQs + Solutions

CUET Mathematics 2024

Calculus

Integrals

Medium

The value of the integral $\int_{\log _{e} 2}^{\log _{e} 3} \frac{e^{2 x}-1}{e^{2 x}+1} d x$ is :

\log _{e} 3

\log _{e} 4-\log _{e} 3

\log _{e} 9-\log _{e} 4

\log _{e} 3-\log _{e} 2

✅ Correct Option: 2

Simplify:

\dfrac{e^{2x}-1}{e^{2x}+1} = \dfrac{e^{2x}+1-2}{e^{2x}+1} = 1 - \dfrac{2}{e^{2x}+1}

So our integral becomes:

\int_{\log_e 2}^{\log_e 3} \left(1 - \dfrac{2}{e^{2x}+1}\right) dx = \int_{\log_e 2}^{\log_e 3} dx - 2\int_{\log_e 2}^{\log_e 3} \dfrac{1}{e^{2x}+1} dx

The first part is straightforward:

\int_{\log_e 2}^{\log_e 3} dx = x\big|_{\log_e 2}^{\log_e 3} = \log_e 3 - \log_e 2 = \log_e \dfrac{3}{2}

For the second part, we use substitution

u = e^{2x}

, giving:

du = 2e^{2x}dx

and

dx = \dfrac{du}{2u}

When

x = \log_e 2

u = e^{2\log_e 2} = 4

\newline

When

x = \log_e 3

u = e^{2\log_e 3} = 9

The integral becomes:

\int_4^9 \dfrac{1}{u+1} \cdot \dfrac{du}{2u} = \dfrac{1}{2}\int_4^9 \dfrac{1}{u(u+1)} du

Using partial fractions:

\dfrac{1}{u(u+1)} = \dfrac{1}{u} - \dfrac{1}{u+1}

Therefore:

\dfrac{1}{2}\int_4^9 \dfrac{1}{u(u+1)} du = \dfrac{1}{2}\int_4^9 \left(\dfrac{1}{u} - \dfrac{1}{u+1}\right) du

= \dfrac{1}{2}\left[\ln|u| - \ln|u+1|\right]_4^9

= \dfrac{1}{2}\left[\ln\left|\dfrac{u}{u+1}\right|\right]_4^9

= \dfrac{1}{2}\left[\ln\left|\dfrac{9}{10}\right| - \ln\left|\dfrac{4}{5}\right|\right]

= \dfrac{1}{2}\ln\left|\dfrac{9/10}{4/5}\right| = \dfrac{1}{2}\ln\left|\dfrac{9 \cdot 5}{10 \cdot 4}\right| = \dfrac{1}{2}\ln\left|\dfrac{45}{40}\right| = \dfrac{1}{2}\ln\left|\dfrac{9}{8}\right|

Combining the results:

\int_{\log_e 2}^{\log_e 3} \dfrac{e^{2x}-1}{e^{2x}+1} dx = \log_e \dfrac{3}{2} - \ln\dfrac{9}{8}

= \ln\dfrac{3}{2} - \ln\dfrac{9}{8} = \ln\left(\dfrac{3/2}{9/8}\right) = \ln\left(\dfrac{3 \cdot 8}{2 \cdot 9}\right) = \ln\left(\dfrac{24}{18}\right) = \ln\left(\dfrac{4}{3}\right)

The value of the integral is

\ln\left(\dfrac{4}{3}\right) = \log _{e} 4-\log _{e} 3