Question

Evaluate: \[ I = \int \frac{5e^x}{(e^x + 1)(e^{2x} + 9)} \, dx. \]

Hint:

For integrals involving rational functions with exponential terms, a substitution like \( t = e^x \) can simplify the expression. Partial fractions can then be used to decompose the integrand.

Answer 1

Step 1: Use Substitution
Let \( t = e^x \), so that \( dt = e^x dx \). Hence, we have: \[ dx = \frac{dt}{t}. \] Now, substitute into the integral: \[ I = \int \frac{5e^x}{(e^x + 1)(e^{2x} + 9)} \, dx = \int \frac{5t}{(t + 1)(t^2 + 9)} \cdot \frac{dt}{t}. \] Simplifying, we get: \[ I = \int \frac{5}{(t + 1)(t^2 + 9)} \, dt. \] Step 2: Decompose Using Partial Fractions
We need to decompose the integrand using partial fractions: \[ \frac{5}{(t + 1)(t^2 + 9)} = \frac{A}{t + 1} + \frac{Bt + C}{t^2 + 9}. \] To solve for \( A \), \( B \), and \( C \), multiply both sides by \( (t + 1)(t^2 + 9) \) to get: \[ 5 = A(t^2 + 9) + (Bt + C)(t + 1). \] Expanding the right-hand side: \[ 5 = A(t^2 + 9) + Bt(t + 1) + C(t + 1), \] \[ 5 = A(t^2 + 9) + Bt^2 + Bt + Ct + C. \] Now, group the terms based on powers of \( t \): \[ 5 = (A + B)t^2 + (B + C)t + (9A + C). \] Equate the coefficients of like terms: - For \( t^2 \): \( A + B = 0 \), - For \( t \): \( B + C = 0 \), - For the constant: \( 9A + C = 5 \). Solving this system of equations: 1. \( A + B = 0 \) gives \( B = -A \), 2. \( B + C = 0 \) gives \( C = -B = A \), 3. Substituting \( B = -A \) and \( C = A \) into \( 9A + C = 5 \): \[ 9A + A = 5 \quad \Rightarrow \quad 10A = 5 \quad \Rightarrow \quad A = \frac{1}{2}. \] Hence, \( B = -\frac{1}{2} \) and \( C = \frac{1}{2} \). Step 3: Substitute the Values of \( A \), \( B \), and \( C \)
Substituting these values into the partial fraction decomposition: \[ \frac{5}{(t + 1)(t^2 + 9)} = \frac{1/2}{t + 1} + \frac{-\frac{1}{2}t + \frac{1}{2}}{t^2 + 9}. \] Thus, we can rewrite the integral as: \[ I = \int \left( \frac{1/2}{t + 1} + \frac{-\frac{1}{2}t + \frac{1}{2}}{t^2 + 9} \right) dt. \] Step 4: Break the Integral Into Separate Terms
Now, split the integral into two parts: \[ I = \frac{1}{2} \int \frac{dt}{t + 1} - \frac{1}{2} \int \frac{t \, dt}{t^2 + 9} + \frac{1}{2} \int \frac{dt}{t^2 + 9}. \] Step 5: Evaluate Each Integral
1. \( \int \frac{dt}{t + 1} = \ln|t + 1| \), 2. \( \int \frac{t \, dt}{t^2 + 9} = \frac{1}{2} \ln|t^2 + 9| \) (standard result), 3. \( \int \frac{dt}{t^2 + 9} = \frac{1}{3} \tan^{-1}\left(\frac{t}{3}\right) \) (standard result). So, the total integral becomes: \[ I = \frac{1}{2} \ln|t + 1| - \frac{1}{4} \ln|t^2 + 9| + \frac{1}{6} \tan^{-1}\left(\frac{t}{3}\right) + C. \] Step 6: Substitute \( t = e^x \) Back Into the Solution
Finally, substitute \( t = e^x \) back into the result: \[ I = \frac{1}{2} \ln|e^x + 1| - \frac{1}{4} \ln|e^{2x} + 9| + \frac{1}{6} \tan^{-1}\left(\frac{e^x}{3}\right) + C. \]