Q.Integrate the following function: $e^{2x+3}$

The key idea is that the integral of $e^{ax+b}$ follows directly from the reverse chain rule: $\int e^{ax+b}\,dx = \frac{1}{a}e^{ax+b} + C$.

Step 1: Identify the exponent. Here, $2x+3$ is a linear function, so $a = 2$ and $b = 3$. …

Why This Works: The Core Idea

When you integrate an exponential function like $e^{f(x)}$, the natural instinct is to think of the derivative rule: $\frac{d}{dx} e^{f(x)} = f'(x) e^{f(x)}$. Integration is the reverse of differentiation, so if you see $e^{f(x)}$, you want to "undo" the chain rule. The catch is that the derivative of $f(x)$ must be present as a factor — or at least a constant multiple — for the integral to be straightforward.

Here, $f(x) = 2x+3$. Its derivative is $f'(x) = 2$, a constant. That means the integrand $e^{2x+3}$ is almost the derivative of $e^{2x+3}$ itself, except it's missing the factor 2. So we compensate by dividing by 2.

Step-by-Step Solution

1. Recognize the form

   The integrand is $e^{2x+3}$. This is an exponential function with a linear exponent. The derivative of the exponent $2x+3$ is $2$, a constant. This signals that the integral will involve a simple adjustment by the reciprocal of that constant.

2. Set up a substitution (optional but clear)

   Let $u = 2x+3$. Then $du = 2 \, dx$, so $dx = \frac{du}{2}$.

   The integral becomes:

3. Integrate the basic exponential …

Method: Integrate an exponential with a linear exponent

Use this for $e^{ax+b}$: the exponential integrates to itself, but the chain rule forces a $\dfrac1a$ correction from the linear exponent.

Steps

Step 1: Recall the base rule and the linear-argument adjustment.

The $\dfrac1a$ appears because differentiating $e^{ax+b}$ brings a factor $a$ out — integration must undo it.

Step 2: Read off $a$ from the exponent. …

Common Mistakes

Mistake 1: Writing $\int e^{2x+3}\,dx=e^{2x+3}$ with no $\dfrac12$.

Why it's wrong: the chain rule means differentiating $e^{2x+3}$ gives $2e^{2x+3}$, so integration must divide by $2$. Correct approach: $\int e^{ax+b}\,dx=\dfrac1a e^{ax+b}+C$.

Mistake 2: Multiplying by $2$ instead of dividing. …