To solve the integral of the function $\frac{x+4}{x+1}$, using the method of partial fractions decomposition (often referred to as the QX/PX formula), we need to follow a few steps.

Step 1: Write the integral in a suitable form

The integral you're asking for is:

$\int \frac{x+4}{x+1} \, dx$

First, notice that the degree of the numerator is equal to the degree of the denominator, so we should first perform polynomial long division to simplify the expression.

Step 2: Perform Polynomial Long Division

We need to divide $x+4$ by $x+1$:

1. Divide the first term of the numerator $x$ by the first term of the denominator $x$: $\frac{x}{x} = 1$.
2. Multiply the result $1$ by the denominator $x+1$ to get $x+1$.
3. Subtract $(x+1)$ from $(x+4)$:

$(x+4) - (x+1) = 3$

So, the quotient is $1$ and the remainder is $3$.

This means that:

$\frac{x+4}{x+1} = 1 + \frac{3}{x+1}$

Step 3: Integrate the Result

Now, we can write the original integral as:

$\int \frac{x+4}{x+1} \, dx = \int 1 \, dx + \int \frac{3}{x+1} \, dx$

The first part is straightforward:

$\int 1 \, dx = x$

For the second part, we use the fact that $\int \frac{1}{x+1} \, dx = \ln|x+1|$:

$\int \frac{3}{x+1} \, dx = 3 \ln|x+1|$

Final Answer

Thus, the solution to the integral is:

$\int \frac{x+4}{x+1} \, dx = x + 3 \ln|x+1| + C$

Where $C$ is the constant of integration.

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Would you like further details on any specific part of the process?

Here are some related questions for you:

1. What happens if the numerator has a higher degree than the denominator?
2. How do you apply the method of partial fractions when the denominator has multiple factors?
3. Can the decomposition method be applied to trigonometric integrals?
4. What’s the role of polynomial long division in partial fraction decomposition?
5. How do you handle repeated factors in the denominator when performing decomposition?

Tip: Always check the degree of the numerator before attempting partial fraction decomposition; if it's larger or equal to the denominator, perform polynomial long division first.