Use the partial fractions to evaluate the integral
Hint: First use the method of substitution with .

Question

Use the partial fractions to evaluate the integral Hint: First use the method of substitution with $$ u=e^{x} $$.

UpStudy AI · Accepted Answer

Certainly! Let's evaluate the integral using the method of substitution followed by partial fractions.

### **Problem:**
Evaluate the integral
$$
\int \frac{e^x}{e^{2x} + e^x - 2} \, dx
$$
**Hint:** First use the substitution $$ u = e^x $$.

### **Solution:**

1. **Substitution:**
   
   Let’s begin by substituting $$ u = e^x $$. Then, the derivative of $$ u $$ with respect to $$ x $$ is:
   $$
   \frac{du}{dx} = e^x \quad \Rightarrow \quad du = e^x \, dx \quad \Rightarrow \quad dx = \frac{du}{e^x} = \frac{du}{u}
   $$
   
   Substitute $$ u $$ and $$ dx $$ into the integral:
   $$
   \int \frac{e^x}{e^{2x} + e^x - 2} \, dx = \int \frac{u}{u^2 + u - 2} \cdot \frac{du}{u} = \int \frac{1}{u^2 + u - 2} \, du
   $$

2. **Partial Fraction Decomposition:**
   
   First, factor the denominator:
   $$
   u^2 + u - 2 = (u + 2)(u - 1)
   $$
   
   Now, express the integrand as partial fractions:
   $$
   \frac{1}{(u + 2)(u - 1)} = \frac{A}{u + 2} + \frac{B}{u - 1}
   $$
   
   To find constants $$ A $$ and $$ B $$, multiply both sides by $$ (u + 2)(u - 1) $$:
   $$
   1 = A(u - 1) + B(u + 2)
   $$
   
   **Solving for $$ A $$ and $$ B $$:**
   
   - Let $$ u = 1 $$:
     $$
     1 = A(1 - 1) + B(1 + 2) \Rightarrow 1 = 3B \Rightarrow B = \frac{1}{3}
     $$
   
   - Let $$ u = -2 $$:
     $$
     1 = A(-2 - 1) + B(-2 + 2) \Rightarrow 1 = -3A \Rightarrow A = -\frac{1}{3}
     $$
   
   So, the partial fractions are:
   $$
   \frac{1}{(u + 2)(u - 1)} = -\frac{1}{3} \cdot \frac{1}{u + 2} + \frac{1}{3} \cdot \frac{1}{u - 1}
   $$

3. **Integrate:**
   
   Now, integrate term by term:
   $$
   \int \frac{1}{u^2 + u - 2} \, du = -\frac{1}{3} \int \frac{1}{u + 2} \, du + \frac{1}{3} \int \frac{1}{u - 1} \, du
   $$
   
   $$
   = -\frac{1}{3} \ln|u + 2| + \frac{1}{3} \ln|u - 1| + C
   $$
   
   Combine the logarithms:
   $$
   = \frac{1}{3} \ln|u - 1| - \frac{1}{3} \ln|u + 2| + C = \frac{1}{3} \ln\left|\frac{u - 1}{u + 2}\right| + C
   $$

4. **Back-Substitute $$ u = e^x $$:**
   
   Replace $$ u $$ with $$ e^x $$:
   $$
   = \frac{1}{3} \ln\left|\frac{e^x - 1}{e^x + 2}\right| + C
   $$

### **Final Answer:**
$$
\int \frac{e^x}{e^{2x} + e^x - 2} \, dx = \frac{1}{3} \ln\left|\frac{e^x - 1}{e^x + 2}\right| + C
$$

where $$ C $$ is the constant of integration.
$$ \int \frac{e^x}{e^{2x} + e^x - 2} \, dx = \frac{1}{3} \ln\left|\frac{e^x - 1}{e^x + 2}\right| + C $$

Use the partial fractions to evaluate the integral
Hint: First use the method of substitution with .

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions

Use the partial fractions to evaluate the integral Hint: First use the method of substitution with .

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions

Use the partial fractions to evaluate the integral
Hint: First use the method of substitution with .