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 \(K+L=\)

• A. \(-\dfrac{3}{38}\)
• B. \(\dfrac{21}{38}\)
• C. \(\dfrac{38}{21}\)
• D. \(-\dfrac{38}{3}\)

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Rewrite the integrand using a different substitution.
Multiply numerator and denominator by e^x: ∫ (3e^(2x) – 7)/(7e^(2x) + 3) dx. Let u = e^(2x), so du = 2e^(2x)dx → dx = du/(2u).

Step 2: Express in terms of u.
Integral becomes ∫ (3u – 7)/(7u + 3) · du/(2u) = (1/2)∫ (3u – 7)/[u(7u+3)] du. Split by partial fractions: (3u–7)/[u(7u+3)] = A/u + B/(7u+3). Solving: A = –7/3, B = 58/3.

Step 3: Integrate and back-substitute.
Integral = (1/2)[ (–7/3)ln|u| + (58/21)ln|7u+3| ] + C. Substitute u = e^(2x): result matches the form Kx + L log(e^(–2x) + 7/3). Comparing gives K = 3/7, L = 29/21.

Step 4: Compute K + L.
K + L = 3/7 + 29/21 = 9/21 + 29/21 = 38/21.

Step 5: Final Answer:
38/21.