To solve the integral [tex]\(\int \frac{x^2 + x + 3}{x^2 + 3} \, dx\)[/tex], we'll proceed with the following steps:
1. Simplify the Integrand: Split the fraction into simpler terms.
[tex]\[
\frac{x^2 + x + 3}{x^2 + 3} = \frac{x^2}{x^2 + 3} + \frac{x}{x^2 + 3} + \frac{3}{x^2 + 3}
\][/tex]
2. Rewrite and Integrate Each Term Separately: We'll deal with each term individually.
- For the first term, [tex]\(\frac{x^2}{x^2 + 3}\)[/tex]:
[tex]\[
\frac{x^2}{x^2 + 3} = 1 - \frac{3}{x^2 + 3}
\][/tex]
This simplifies our integral into:
[tex]\[
\int \left(1 - \frac{3}{x^2 + 3} + \frac{x}{x^2 + 3} + \frac{3}{x^2 + 3}\right) \, dx
\][/tex]
Simplifying further, we get:
[tex]\[
\int 1 \, dx + \int \frac{x}{x^2 + 3} \, dx
\][/tex]
3. Integrate the Simplified Form:
- The integral of a constant 1:
[tex]\[
\int 1 \, dx = x
\][/tex]
- For the term [tex]\(\int \frac{x}{x^2 + 3} \, dx\)[/tex], let [tex]\(u = x^2 + 3\)[/tex]. Then [tex]\(du = 2x \, dx\)[/tex], and [tex]\(dx = \frac{du}{2x}\)[/tex]:
[tex]\[
\int \frac{x}{x^2 + 3} \, dx = \int \frac{1}{2} \cdot \frac{du}{u} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u| = \frac{1}{2} \ln|x^2 + 3|
\][/tex]
4. Combine the Results:
Summing up the integrated parts, the complete integral is:
[tex]\[
x + \frac{1}{2} \ln|x^2 + 3| + C
\][/tex]
Thus, the result of the integral [tex]\(\int \frac{x^2 + x + 3}{x^2 + 3} \, dx\)[/tex] is:
[tex]\[
x + \frac{1}{2} \ln(x^2 + 3) + C
\][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
