Question:
If \(\displaystyle \int \frac{3e^x-7e^{-x}}{7e^x+3e^{-x}}\,dx=Kx+L\log\left(e^{-2x}+\frac{7}{3}\right)+C\), then \(K+L=\)
If \(\displaystyle \int \frac{3e^x-7e^{-x}}{7e^x+3e^{-x}}\,dx=Kx+L\log\left(e^{-2x}+\frac{7}{3}\right)+C\), then \(K+L=\)
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When an integrand contains both \(e^x\) and \(e^{-x}\), divide numerator and denominator by \(e^x\) to convert it into an expression involving \(e^{-2x}\). This often makes comparison with logarithmic forms easier.
Updated On: Jun 18, 2026
- \(-\dfrac{3}{38}\)
- \(\dfrac{21}{38}\)
- \(\dfrac{38}{21}\)
- \(-\dfrac{38}{3}\)
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The Correct Option is C
Solution and Explanation
Step 1: Simplify the integrand.
Given integral is \[ \int \frac{3e^x-7e^{-x}}{7e^x+3e^{-x}}\,dx \] Divide numerator and denominator by \(e^x\), we get \[ \int \frac{3-7e^{-2x}}{7+3e^{-2x}}\,dx \] Let \[ t=e^{-2x} \] Then the integrand becomes \[ \frac{3-7t}{7+3t} \]
Step 2: Express the integrand in required form.
We need to write \[ \frac{3-7t}{7+3t} \] in the form \[ K+M\cdot \frac{t}{t+\frac{7}{3}} \] Since \[ 7+3t=3\left(t+\frac{7}{3}\right), \] we have \[ \frac{3-7t}{7+3t} = \frac{1-\frac{7}{3}t}{t+\frac{7}{3}} \] Now compare \[ K+M\frac{t}{t+\frac{7}{3}} = \frac{K\left(t+\frac{7}{3}\right)+Mt}{t+\frac{7}{3}} \] \[ = \frac{(K+M)t+\frac{7K}{3}}{t+\frac{7}{3}} \] Comparing coefficients with \[ \frac{1-\frac{7}{3}t}{t+\frac{7}{3}}, \] we get \[ \frac{7K}{3}=1 \] So, \[ K=\frac{3}{7} \] Also, \[ K+M=-\frac{7}{3} \] Therefore, \[ M=-\frac{7}{3}-\frac{3}{7} \] \[ M=\frac{-49-9}{21} \] \[ M=-\frac{58}{21} \]
Step 3: Relate \(M\) with \(L\).
We know \[ \frac{d}{dx}\log\left(e^{-2x}+\frac{7}{3}\right) = \frac{-2e^{-2x}}{e^{-2x}+\frac{7}{3}} \] So, \[ L\frac{d}{dx}\log\left(e^{-2x}+\frac{7}{3}\right) = -2L\frac{e^{-2x}}{e^{-2x}+\frac{7}{3}} \] Hence, \[ M=-2L \] Using \[ M=-\frac{58}{21}, \] we get \[ -2L=-\frac{58}{21} \] \[ L=\frac{29}{21} \]
Step 4: Find \(K+L\).
\[ K+L=\frac{3}{7}+\frac{29}{21} \] \[ K+L=\frac{9}{21}+\frac{29}{21} \] \[ K+L=\frac{38}{21} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{38}{21}} \]
Given integral is \[ \int \frac{3e^x-7e^{-x}}{7e^x+3e^{-x}}\,dx \] Divide numerator and denominator by \(e^x\), we get \[ \int \frac{3-7e^{-2x}}{7+3e^{-2x}}\,dx \] Let \[ t=e^{-2x} \] Then the integrand becomes \[ \frac{3-7t}{7+3t} \]
Step 2: Express the integrand in required form.
We need to write \[ \frac{3-7t}{7+3t} \] in the form \[ K+M\cdot \frac{t}{t+\frac{7}{3}} \] Since \[ 7+3t=3\left(t+\frac{7}{3}\right), \] we have \[ \frac{3-7t}{7+3t} = \frac{1-\frac{7}{3}t}{t+\frac{7}{3}} \] Now compare \[ K+M\frac{t}{t+\frac{7}{3}} = \frac{K\left(t+\frac{7}{3}\right)+Mt}{t+\frac{7}{3}} \] \[ = \frac{(K+M)t+\frac{7K}{3}}{t+\frac{7}{3}} \] Comparing coefficients with \[ \frac{1-\frac{7}{3}t}{t+\frac{7}{3}}, \] we get \[ \frac{7K}{3}=1 \] So, \[ K=\frac{3}{7} \] Also, \[ K+M=-\frac{7}{3} \] Therefore, \[ M=-\frac{7}{3}-\frac{3}{7} \] \[ M=\frac{-49-9}{21} \] \[ M=-\frac{58}{21} \]
Step 3: Relate \(M\) with \(L\).
We know \[ \frac{d}{dx}\log\left(e^{-2x}+\frac{7}{3}\right) = \frac{-2e^{-2x}}{e^{-2x}+\frac{7}{3}} \] So, \[ L\frac{d}{dx}\log\left(e^{-2x}+\frac{7}{3}\right) = -2L\frac{e^{-2x}}{e^{-2x}+\frac{7}{3}} \] Hence, \[ M=-2L \] Using \[ M=-\frac{58}{21}, \] we get \[ -2L=-\frac{58}{21} \] \[ L=\frac{29}{21} \]
Step 4: Find \(K+L\).
\[ K+L=\frac{3}{7}+\frac{29}{21} \] \[ K+L=\frac{9}{21}+\frac{29}{21} \] \[ K+L=\frac{38}{21} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{38}{21}} \]
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