Answer:
\(\displaystyle \int {\frac{e^{3x}}{e^{6x} + 1}} \, dx = \frac{arctan(e^{3x})}{3} + C\)
General Formulas and Concepts:
Algebra I
- Exponential Rule [Powering]: \(\displaystyle (b^m)^n = b^{m \cdot n}\)
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Rule [Chain Rule]: \(\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)\)
Integration
- Integrals
- Indefinite Integrals
- Integration Constant C
Integration Property [Multiplied Constant]: \(\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx\)
U-Substitution
Step-by-step explanation:
Step 1: Define
Identify
\(\displaystyle \int {\frac{e^{3x}}{e^{6x} + 1}} \, dx\)
Step 2: Integrate Pt. 1
Identify variables for u-substitution.
- Set u: \(\displaystyle u = e^{3x}\)
- [u] Differentiate [Exponential Differentiation, Chain Rule]: \(\displaystyle du = 3e^{3x} \ dx\)
Step 3: Integrate Pt. 2
- [Integrand] Rewrite [Exponential Rule - Powering]: \(\displaystyle \int {\frac{e^{3x}}{e^{6x} + 1}} \, dx = \int {\frac{e^{3x}}{(e^{3x})^2 + 1}} \, dx\)
- [Integral] Rewrite [Integration Property - Multiplied Constant]: \(\displaystyle \int {\frac{e^{3x}}{e^{6x} + 1}} \, dx = \frac{1}{3}\int {\frac{3e^{3x}}{(e^{3x})^2 + 1}} \, dx\)
- [Integral] U-Substitution: \(\displaystyle \int {\frac{e^{3x}}{e^{6x} + 1}} \, dx = \frac{1}{3}\int {\frac{1}{u^2 + 1}} \, du\)
- [Integral] Arctrig Integration: \(\displaystyle \int {\frac{e^{3x}}{e^{6x} + 1}} \, dx = \frac{1}{3} \bigg[ \frac{1}{1}arctan \Big( \frac{u}{1} \Big) \bigg] + C\)
- Simplify: \(\displaystyle \int {\frac{e^{3x}}{e^{6x} + 1}} \, dx = \frac{arctan(u)}{3} + C\)
- Back-Substitute: \(\displaystyle \int {\frac{e^{3x}}{e^{6x} + 1}} \, dx = \frac{arctan(e^{3x})}{3} + C\)
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
