Question

Let \[ f(x)=\int \frac{e^{3x}}{4+8e^{2x}+e^{4x}}\,dx \] and \[ g(x)=\int \frac{2\,dx}{e^{3x}+8e^x+4e^{-x}}, \] then \[ f(x)-g(x)= \] is:

• A. \(\dfrac{1}{2}\tan^{-1}\left(\dfrac{e^x+2e^{-x}}{2}\right)+C\)
• B. \(\dfrac{1}{2}\tan^{-1}\left(\dfrac{e^x+e^{-x}}{2}\right)+C\)
• C. \(\dfrac{1}{2}\tan^{-1}\left(\dfrac{2e^{-x}+e^{-2x}}{2}\right)+C\)
• D. \(\dfrac{1}{2}\tan^{-1}\left(\dfrac{e^{2x}+2e^x}{2e^x}\right)+C\)

Hint:

When two integrals have exponential expressions, first rewrite both integrands with a common denominator. Then look for a substitution whose derivative appears in the numerator.

Answer 1

The Correct Option is A

Step 1: Rewrite \(g(x)\) with the same denominator.
Given, \[ g(x)=\int \frac{2\,dx}{e^{3x}+8e^x+4e^{-x}} \] Multiplying numerator and denominator by \(e^x\), we get \[ g(x)=\int \frac{2e^x}{e^{4x}+8e^{2x}+4}\,dx \] Also, \[ f(x)=\int \frac{e^{3x}}{e^{4x}+8e^{2x}+4}\,dx \]

Step 2: Find \(f(x)-g(x)\).
\[ f(x)-g(x) = \int \frac{e^{3x}-2e^x}{e^{4x}+8e^{2x}+4}\,dx \] \[ = \int \frac{e^x(e^{2x}-2)}{e^{4x}+8e^{2x}+4}\,dx \]

Step 3: Identify a suitable substitution.
Let \[ u=\frac{e^x+2e^{-x}}{2} \] Then, \[ \frac{du}{dx} = \frac{e^x-2e^{-x}}{2} \] Also, \[ 1+u^2 = 1+\left(\frac{e^x+2e^{-x}}{2}\right)^2 \] \[ = \frac{e^{2x}+8+4e^{-2x}}{4} \] Now, \[ \frac{1}{2}\cdot \frac{u'}{1+u^2} = \frac{e^{3x}-2e^x}{e^{4x}+8e^{2x}+4} \]

Step 4: Integrate.
Therefore, \[ f(x)-g(x) = \frac{1}{2}\tan^{-1}u+C \] Substitute \[ u=\frac{e^x+2e^{-x}}{2} \] Thus, \[ f(x)-g(x) = \frac{1}{2}\tan^{-1}\left(\frac{e^x+2e^{-x}}{2}\right)+C \]

Step 5: Final conclusion.
Hence, \[ \boxed{ \frac{1}{2}\tan^{-1}\left(\frac{e^x+2e^{-x}}{2}\right)+C } \]