The value of the integral _{ _{e} 2}^{ _{e} 3} {e² x}-1}{e² x}+1} d x is :

Correct Answer is : loge 4 – loge 3

Correct Answer is : loge 4 – loge 3

Test Index

AP DSC School Assistant Math Model Test 3

Section: Subject Specific

Question : 63 of 160

Marks: +1, -0

\rm \displaystyle \int_{\log _{e} 2}^{\log _{e} 3} \frac{e^{2 x}-1}{e^{2 x}+1} d x

is :

Solution:

Concept:

We evaluate the definite integral by substitution and partial fraction decomposition.

Explanation:

Let I =

\int_{\ln 2}^{\ln 3} \frac{e^{2x}-1}{e^{2x}+1}\,dx

Substitute

t = e^{2x}

⇒

dt = 2e^{2x}dx = 2t\,dx

⇒

dx = \frac{dt}{2t}

When

x = \ln 2

t = e^{2\ln 2}=4

; when

x = \ln 3

t = e^{2\ln 3}=9

Then

I = \int_{4}^{9} \frac{t-1}{t+1}\cdot\frac{1}{2t}\,dt = \frac12\int_{4}^{9} \frac{t-1}{t(t+1)}\,dt.

Decompose the integrand into partial fractions:

\frac{t-1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1}.

Multiplying both sides by

t(t+1)

t-1 = A(t+1) + Bt

⇒

t-1 = (A+B)t + A

Comparing coefficients:

A+B=1

and

A=-1

⇒

B=2

Thus

I = \frac12\int_{4}^{9} \left(-\frac1t + \frac{2}{t+1}\right) dt = \frac12\Bigl[-\ln|t| + 2\ln|t+1|\Bigr]_{4}^{9}.

Evaluate:

\begin{aligned} I &= \frac12\Bigl[(-\ln 9 + 2\ln 10) - (-\ln 4 + 2\ln 5)\Bigr] \\ &= \frac12\Bigl[-\ln 9 + 2\ln 10 + \ln 4 - 2\ln 5\Bigr] \\ &= \frac12\Bigl[\ln 4 - \ln 9 + 2(\ln 10 - \ln 5)\Bigr] \\ &= \frac12\Bigl[\ln 4 - \ln 9 + 2\ln 2\Bigr] \quad(\because \ln 10 - \ln 5 = \ln 2) \\ &= \frac12\Bigl[\ln (4\cdot 4) - \ln 9\Bigr] = \frac12\Bigl[\ln 16 - \ln 9\Bigr] \\ &= \frac12\ln\frac{16}{9} = \ln\sqrt{\frac{16}{9}} = \ln\frac{4}{3}. \end{aligned}

Hence the value is

\ln 4 - \ln 3

(or

\log_e 4 - \log_e 3

Answer:

Option B:

\log_e 4 - \log_e 3

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