The integral presented in the image is:

$\int \frac{7x^2 + 3x + 1}{x^2 + x} \, dx$

To solve this, we can break down the expression into simpler components by dividing the numerator by the denominator using polynomial division.

Step-by-Step Solution:

1. Perform Polynomial Division: Divide $7x^2 + 3x + 1$ by $x^2 + x$.

   Start by dividing the leading term of the numerator $7x^2$ by the leading term of the denominator $x^2$, which gives $7$.

   Multiply $7$ by $x^2 + x$, which results in $7x^2 + 7x$.

   Subtract $7x^2 + 7x$ from $7x^2 + 3x + 1$, giving the new remainder $-4x + 1$.

2. Rewrite the Integral: Now we have:

   $\int \left( 7 + \frac{-4x + 1}{x^2 + x} \right) dx$

   Split the integral:

   $\int 7 \, dx + \int \frac{-4x + 1}{x^2 + x} \, dx$

3. Simplify the Second Integral: Factor the denominator $x^2 + x = x(x+1)$, so we have:

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

   Now, decompose the fraction into partial fractions:

   $\frac{-4x + 1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$

   Multiply both sides by $x(x+1)$:

   $-4x + 1 = A(x+1) + Bx$

   Expand and collect terms:

   $-4x + 1 = A(x) + A + Bx$ $-4x + 1 = (A + B)x + A$

   Comparing coefficients: $A + B = -4 \quad \text{and} \quad A = 1$

   Solving these, we get $A = 1$ and $B = -5$.

   Therefore, the decomposition is:

   $\frac{-4x + 1}{x(x+1)} = \frac{1}{x} - \frac{5}{x+1}$

4. Integrate: Now, the integral becomes:

   $7x + \int \frac{1}{x} \, dx - 5 \int \frac{1}{x+1} \, dx$

   The integrals of the logarithmic terms are:

   $7x + \ln|x| - 5 \ln|x+1| + C$

   Where $C$ is the constant of integration.

Final Answer:

$7x + \ln|x| - 5\ln|x+1| + C$

Would you like further details on any part of the solution, or do you have any questions?

Related Questions:

1. How is polynomial division used in integration?
2. What is the method of partial fractions, and when is it applied?
3. How do logarithmic terms arise from integrals of rational functions?
4. Can substitution methods be used in this integral instead of partial fractions?
5. How do we determine the constant of integration $C$ in definite integrals?

Tip: When facing complex rational integrals, partial fractions often simplify the process by breaking down fractions into simpler terms.