Question

If \(\int \frac {2e^x + e^x}{3e^x + 4e^{-x}} \,dx\) = Ax + Blog( 3e^2x + 4) + C, then values of A and B are respectively (where C is a constant of integration.)

• A. \(\frac {3}{4}, \frac {1}{24}\)
• B. \(\frac {4}{3}, - 24\)
• C. \(\frac {1}{4}, \frac {1}{24}\)
• D. \(\frac {3}{4}, -\frac {1}{24}\)

Answer 1

The Correct Option is D

\(\frac {2e^x + 3e^{–x}}{3e^x + 4e^{–x}}\) = \(\frac {2e^{2x} + 3}{3e^{2x} + 4}\)
2e^2x + 3 = A(3e^2x + 4) + B(6 ∙ e^2x)
2 = 3A + 6B & 3 = 4A ⇒ A =\(\frac {3}{4}\)
6B = 2 – 3A 
6B = 2 – \(\frac {9}{4}\) 
6B = \(-\frac {1}{4}\)
B =  \(-\frac {1}{24}\)
I = ∫[(2e2x + 3)/(3e2x + 4)]dx.
I = ∫[({+ \(\frac {3}{4}\))(3e2x + 4) \(-\frac {1}{24}\)\(\frac {6 \times e^{2x}dx}{(3e^{2x} + 4)dx}\)dx.
I = \(\frac {3}{4}\)∫dx \(-\frac {1}{24}\))∫[(6 ∙ e2x)/(3e2x + 4)]dx.
I = \(\frac {3}{4}\)x \(-\frac {1}{24}\) log |3e^2x + 4| + c;
by comparing,
 A = \(\frac {3}{4}\) and B = – \(\frac {1}{24}\)