To solve the definite integral [tex]\(\int_0^2 \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx\)[/tex], we follow these steps:
1. Identify the Form: The integrand [tex]\(\frac{2x + 1}{\sqrt{x^2 + 4}}\)[/tex] suggests that we may need to use substitution to simplify the expression.
2. Substitution: Consider the substitution [tex]\(u = x^2 + 4\)[/tex]. Then, [tex]\(du = 2x \, dx\)[/tex]. This transforms parts of the integrand accordingly.
3. Rewriting Integral: When [tex]\(x = 0\)[/tex], [tex]\(u = 4\)[/tex]. When [tex]\(x = 2\)[/tex], [tex]\(u = 8\)[/tex].
[tex]\[
\int_{4}^{8} \frac{2x}{\sqrt{u}} \frac{du}{2x} + \int_{4}^{8} \frac{1}{\sqrt{u}} du
\][/tex]
Simplifying within the integral context, it separates as:
[tex]\[
\int_{4}^{8} \frac{1}{\sqrt{u}} du + \int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} dx
\][/tex]
4. Solving Each Integral:
- First integral: [tex]\(\int_{4}^{8} \frac{1}{\sqrt{u}} \, du\)[/tex]
[tex]\[
\int_{4}^{8} u^{-1/2} \, du = 2 u^{1/2} \Big|_{4}^{8}
\][/tex]
Calculating this result:
[tex]\[
= 2\sqrt{8} - 2\sqrt{4} = 4\sqrt{2} - 4
\][/tex]
- Second integral: [tex]\(\int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} \, dx\)[/tex]
For this, set [tex]\(u = x^2 + 4\)[/tex] and then:
[tex]\[
= \int_{4}^{8} u^{-1/2} \cdot \frac{du}{2x} + \int_{0}^{2} \frac{1}{\sqrt{x^2+4}} dx = \log(1 + \sqrt{2})
\][/tex]
5. Combining Results: The results from both solved integrals sum up:
[tex]\[
\int_4^8 \frac{1}{\sqrt{u}} \, du + \int_{0}^{2} \frac{1}{\sqrt{x^2+4}}dx = -4 + log(1 + \sqrt(2)) + 4\sqrt{2}
\][/tex]
Thus, the value of the definite integral [tex]\(\int_0^2 \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx\)[/tex] is [tex]\[-4 + \log(1 + \sqrt{2}) + 4\sqrt{2}\][/tex].
