Integrating Exponentials & Logarithms

Since $\frac{d}{dx}(e^x) = e^x$, it follows immediately that $\int e^x\,dx = e^x + C$. For $e^{kx}$, we use the chain rule in reverse: differentiation multiplied by $k$, so integration must divide by $k$.

Verification: $\frac{d}{dx}\!\left(\frac{1}{k}e^{kx}\right) = \frac{1}{k} \cdot k \cdot e^{kx} = e^{kx}$ ✓

Examples:

• $\int e^{3x}\,dx = \dfrac{1}{3}e^{3x} + C$
• $\int e^{-2x}\,dx = -\dfrac{1}{2}e^{-2x} + C$
• $\int 5e^{0.1x}\,dx = 5 \cdot \dfrac{1}{0.1}e^{0.1x} + C = 50e^{0.1x} + C$

$\int e^x\,dx = e^x + C$, so $e^x$ reproduces itself under integration up to the constant; $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$, divide by the inner constant $k$

Pause, copy the rules $\int e^x\,dx = e^x + C$ (an indefinite integral is a FAMILY, so never drop the $+C$) and $\int e^{kx}\,dx = \dfrac{1}{k}e^{kx} + C$ (divide by the inner constant $k$) into your book.

We just saw that $\int e^{kx}\,dx = \dfrac{1}{k}e^{kx} + C$ by reversing the chain rule. That raises a question: the power rule says $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$, but what happens when $n = -1$, making the denominator zero? This card answers it → the power rule breaks down there, and $\int x^{-1}\,dx = \ln|x| + C$ instead.

The power rule $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ fails for $n = -1$ because the denominator becomes zero. But we know $\frac{d}{dx}(\ln x) = \frac{1}{x}$ for $x > 0$. For $x < 0$, $\frac{d}{dx}(\ln(-x)) = \frac{-1}{-x} = \frac{1}{x}$. Combining both cases:

The absolute value is essential. $\ln x$ is only defined for $x > 0$, but $\frac{1}{x}$ exists for all $x \ne 0$. The absolute value lets us cover both positive and negative domains.

$\int \frac{1}{x}\,dx = \ln|x| + C$, this is the only exception to the power rule; The power rule fails at $n = -1$ because $\frac{x^0}{0}$ is undefined (division by zero)

Pause, copy the rule $\int \dfrac{1}{x}\,dx = \ln|x| + C$, the only exception to the power rule, necessary because $\dfrac{x^0}{0}$ is undefined, into your book.

We just saw that $\int \frac{1}{x}\,dx = \ln|x| + C$ because the power rule fails at $n = -1$. That raises a question: what about $\int 2^x\,dx$ or $\int 3^x\,dx$, where the base is not $e$? This card answers it → dividing $a^x \ln a$ (the derivative of $a^x$) by $\ln a$ gives $\int a^x\,dx = \dfrac{a^x}{\ln a} + C$.

Since $\frac{d}{dx}(a^x) = a^x \ln a$, dividing both sides by $\ln a$ gives the integral rule:

Verification: $\frac{d}{dx}\!\left(\frac{a^x}{\ln a}\right) = \frac{a^x \ln a}{\ln a} = a^x$ ✓

Example: $\int 2^x\,dx = \dfrac{2^x}{\ln 2} + C$.

This rule is less common in HSC than $e^x$ because any $a^x$ can be rewritten as $e^{x \ln a}$. However, if you see $2^x$ or $3^x$, apply this formula directly, it's cleaner than converting.

$\int a^x\,dx = \frac{a^x}{\ln a} + C$ where $a > 0, a \ne 1$; When $a = e$: $\frac{e^x}{\ln e} = \frac{e^x}{1} = e^x$, confirms $\int e^x\,dx = e^x + C$

Pause, copy the rule $\int a^x\,dx = \dfrac{a^x}{\ln a} + C$ and verify it reduces to $\int e^x\,dx = e^x + C$ when $a = e$ (since $\ln e = 1$) into your book.