Question

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

Answer: (D)

(D)

Answer 2

Let the given integral be $I$.

$I = \int \frac{2e^x + 3e^{-x}}{3e^x + 4e^{-x}} dx$

We are given that the result of the integral is $Ax + Blog(3e^{2x} + 4) + C$.

Let $F(x) = Ax + Blog(3e^{2x} + 4) + C$.

Then, the derivative of $F(x)$ with respect to $x$ must be equal to the integrand.

$F'(x) = \frac{d}{dx} (Ax + Blog(3e^{2x} + 4) + C)$

$F'(x) = A + B \frac{d}{dx}(log(3e^{2x} + 4)) + 0$

Using the chain rule, $\frac{d}{dx}(log(u)) = \frac{1}{u} \frac{du}{dx}$. Let $u = 3e^{2x} + 4$.

$\frac{du}{dx} = \frac{d}{dx}(3e^{2x} + 4) = 3 \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(4) = 3(e^{2x} \cdot 2) + 0 = 6e^{2x}$.

So, $\frac{d}{dx}(log(3e^{2x} + 4)) = \frac{6e^{2x}}{3e^{2x} + 4}$.

Thus, $F'(x) = A + B \frac{6e^{2x}}{3e^{2x} + 4}$.

We need to equate this to the integrand:

$\frac{2e^x + 3e^{-x}}{3e^x + 4e^{-x}} = A + B \frac{6e^{2x}}{3e^{2x} + 4}$

Let's simplify the integrand by multiplying the numerator and the denominator by $e^x$:

$\frac{(2e^x + 3e^{-x})e^x}{(3e^x + 4e^{-x})e^x} = \frac{2e^{2x} + 3e^0}{3e^{2x} + 4e^0} = \frac{2e^{2x} + 3}{3e^{2x} + 4}$.

Now, equate the simplified integrand to $F'(x)$:

$\frac{2e^{2x} + 3}{3e^{2x} + 4} = A + B \frac{6e^{2x}}{3e^{2x} + 4}$

Combine the terms on the right side by finding a common denominator:

$A + B \frac{6e^{2x}}{3e^{2x} + 4} = \frac{A(3e^{2x} + 4)}{3e^{2x} + 4} + \frac{6Be^{2x}}{3e^{2x} + 4} = \frac{3Ae^{2x} + 4A + 6Be^{2x}}{3e^{2x} + 4}$

$= \frac{(3A + 6B)e^{2x} + 4A}{3e^{2x} + 4}$.

Equating the numerators:

$2e^{2x} + 3 = (3A + 6B)e^{2x} + 4A$.

This equation must hold for all values of $x$. By comparing the coefficients of $e^{2x}$ and the constant terms on both sides, we get a system of linear equations for A and B:

Comparing coefficients of $e^{2x}$: $2 = 3A + 6B$

Comparing constant terms: $3 = 4A$

From the second equation, $4A = 3 \implies A = \frac{3}{4}$.

Substitute the value of A into the first equation:

$2 = 3\left(\frac{3}{4}\right) + 6B$

$2 = \frac{9}{4} + 6B$

$6B = 2 - \frac{9}{4}$

$6B = \frac{8}{4} - \frac{9}{4}$

$6B = -\frac{1}{4}$

$B = -\frac{1}{4 \times 6}$

$B = -\frac{1}{24}$.

Thus, the values of A and B are $\frac{3}{4}$ and $-\frac{1}{24}$ respectively.