Question

If $\int \frac {2e^x + e^x}{3e^x + 4e^{-x}} \,dx$ = Ax + Blog( 3e2x + 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

Answer: (D)

$\frac {3}{4}, -\frac {1}{24}$

Answer 2

$\frac {2e^x + 3e^{–x}}{3e^x + 4e^{–x}}$ = $\frac {2e^{2x} + 3}{3e^{2x} + 4}$
2e2x + 3 = A(3e2x + 4) + B(6 ∙ e2x)
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 |3e2x + 4| + c;
by comparing,
A = $\frac {3}{4}$ and B = – $\frac {1}{24}$