Integrating Exponential Functions

Integrating Exponential Functions

Basic Antiderivative

• The antiderivative of $e^x$ is simply $e^x + C$.

Antiderivative with Constant Multiple in Exponent

• The antiderivative of $e^{2x}$ is $\frac{e^{2x}}{2} + C$, where 2 is the derivative of $2x$.
• This technique works only when dividing by a constant, not a variable.

Example

• The antiderivative of $e^{5x}$ is $\frac{e^{5x}}{5} + C$.

Proof using u-substitution

• Let $u = 5x$. Then $\frac{du}{dx} = 5$, so $du = 5 dx$.
• Solving for $dx$, we get $dx = \frac{du}{5}$.
• Substituting into the integral, we have $\int e^{5x} dx = \int e^u \frac{du}{5} = \frac{1}{5} \int e^u du$.
• The antiderivative of $e^u$ is $e^u + C$, so we have $\frac{1}{5}e^u + C$.
• Substituting back $u = 5x$, we get $\frac{1}{5} e^{5x} + C$.

Example with Substitution

• Integrate $\int x^3 e^{x^4} dx$.
• Let $u = x^4$. Then $\frac{du}{dx} = 4x^3$, so $du = 4x^3 dx$.
• Solving for $dx$, we get $dx = \frac{du}{4x^3}$.
• Substituting, we have $\int x^3 e^{x^4} dx = \int x^3 e^u \frac{du}{4x^3} = \frac{1}{4} \int e^u du$.
• The antiderivative of $e^u$ is $e^u + C$, so we have $\frac{1}{4} e^u + C$.
• Substituting back $u = x^4$, we get $\frac{1}{4} e^{x^4} + C$.

Another Example with Substitution

• Integrate $\int e^x \sqrt{1 - e^x} dx$.
• Let $u = 1 - e^x$. Then $\frac{du}{dx} = -e^x$, so $du = -e^x dx$.
• Solving for $dx$, we get $dx = \frac{du}{-e^x}$.
• Substituting, we have $\int e^x \sqrt{1 - e^x} dx = \int e^x \sqrt{u} \frac{du}{-e^x} = -\int \sqrt{u} du = -\int u^{\frac{1}{2}} du$.
• Integrating, we get $- \frac{2}{3} u^{\frac{3}{2}} + C$.
• Substituting back $u = 1 - e^x$, we get $\frac{-2}{3}(1 - e^x)^{\frac{3}{2}} + C$.

Example with Natural Logarithm

• Integrate $\int \frac{e^x + e}{e^x - e} dx$.
• Let $u = e^x - e$. Then $\frac{du}{dx} = e^x + e$, so $du = (e^x + e) dx$.
• Solving for $dx$, we get $dx = \frac{du}{e^x + e}$.
• Substituting, we have $\int \frac{e^x + e}{e^x - e} dx = \int \frac{e^x + e}{u} \frac{du}{e^x + e} = \int \frac{1}{u} du$.
• The antiderivative of $\frac{1}{u}$ is $\ln|u| + C$, so we have $\ln|u| + C$.
• Substituting back $u = e^x - e$, we get $\ln|e^x - e| + C$.

Example with Negative Exponents

• Integrate $\int \frac{e^{\frac{1}{x^2}}}{x^3} dx$.
• Let $u = \frac{1}{x^2} = x^{-2}$. Then $\frac{du}{dx} = -2x^{-3} = \frac{-2}{x^3}$, so $du = \frac{-2}{x^3} dx$.
• Solving for $dx$, we get $dx = \frac{x^3}{-2} du$.
• Substituting, we have $\int \frac{e^{\frac{1}{x^2}}}{x^3} dx = \int \frac{e^u}{x^3} \frac{x^3}{-2} du = \frac{-1}{2} \int e^u du$.
• The antiderivative of $e^u$ is $e^u + C$, so we have $\frac{-1}{2} e^u + C$.
• Substituting back $u = \frac{1}{x^2}$, we get $\frac{-1}{2} e^{\frac{1}{x^2}} + C$.

Splitting Fractions

• Integrate $\int \frac{e^{3x} + 4e^x + 5}{e^x} dx$.
• Split the fraction: $\int \frac{e^{3x}}{e^x} + \frac{4e^x}{e^x} + \frac{5}{e^x} dx = \int e^{2x} + 4 + 5e^{-x} dx$.
• Using the rule $\frac{x^a}{x^b} = x^{a-b}$, we subtract the exponents.
• The antiderivative of $e^{2x}$ is $\frac{e^{2x}}{2}$.
• The antiderivative of 4 is $4x$.
• The antiderivative of $5e^{-x}$ is $\frac{5e^{-x}}{-1} = -5e^{-x}$.
• So the final answer is $\frac{e^{2x}}{2} + 4x - 5e^{-x} + C$.