Q.$\int \dfrac{x+3}{(x+4)^2}\,e^{x}\,dx =$ _______.

The key idea is to rewrite the integrand so that the numerator matches the derivative of the denominator, enabling the use of the standard result $\int e^x \bigl(f(x) + f'(x)\bigr)\,dx = e^x f(x) + C$.

First, write the numerator as:

Then the integral becomes:

$\int \frac{(x+4) - 1}{(x+4)^2}\,e^x\,dx = \int \left( \frac{1}{x+4} - \frac{1}{(x+4)^2} \right) e^x\,dx.$ …

We start with the integral:

The presence of $e^x$ strongly suggests using the standard result:

This works because the derivative of $e^x f(x)$ is $e^x f(x) + e^x f'(x) = e^x [f(x) + f'(x)]$.

So our goal is to express $\frac{x+3}{(x+4)^2}$ as $f(x) + f'(x)$ for some function $f(x)$. Let's find that $f(x)$.

1. Look for a candidate $f(x)$. The denominator $(x+4)^2$ suggests $f(x)$ might be of the form $\frac{1}{x+4}$ or $\frac{A}{x+4}$. Let's try $f(x) = \frac{1}{x+4}$. Then:

So:

Perfect! That's exactly our numerator over the denominator.

2. Apply the standard formula. Since $\frac{x+3}{(x+4)^2} = f(x) + f'(x)$ with $f(x) = \frac{1}{x+4}$, we have:

3. Check by differentiating (optional but reassuring). Differentiate $e^x \cdot \frac{1}{x+4}$: …

Method: The $\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+C$ rule

The go-to trick whenever $e^x$ multiplies a rational function: write the rational part as some $f(x)$ plus its own derivative $f'(x)$.

Steps

Step 1: Guess $f$ from the denominator.

A denominator $(x+k)^2$ suggests $f(x)=\dfrac{1}{x+k}$, whose derivative is $-\dfrac{1}{(x+k)^2}$.

Step 2: Verify that $f+f'$ matches. …

Common Mistakes

Mistake 1: Reaching straight for integration by parts.

Why it's wrong: it is far messier than the $e^x[f+f']$ pattern. Correct approach: split $x+3=(x+4)-1$ to reveal $f=\frac{1}{x+4}$ and $f'=-\frac{1}{(x+4)^2}$.

Mistake 2: Mis-differentiating $f=\frac{1}{x+4}$. …