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Certainly! Let's solve the integral [tex]\(\int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx\)[/tex].
First, identify the integrand: [tex]\(\frac{2x + 1}{\sqrt{x^2 + 4}}\)[/tex].
To solve this integral, we can use techniques such as substitution and recognizing standard forms of integrals. Here is a detailed step-by-step solution:
1. Recognize that the expression under the square root [tex]\(x^2 + 4\)[/tex] suggests using a trigonometric or hyperbolic substitution. However, we can also consider a more straightforward method by splitting the integrand.
2. Split the integrand:
[tex]\[ \int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx = \int_{0}^{2} \frac{2x}{\sqrt{x^2 + 4}} \, dx + \int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} \, dx \][/tex]
3. Solve the first part [tex]\(\int_{0}^{2} \frac{2x}{\sqrt{x^2 + 4}} \, dx\)[/tex]:
Let [tex]\(u = x^2 + 4\)[/tex]. Then, [tex]\(du = 2x \, dx\)[/tex].
When [tex]\(x = 0\)[/tex], [tex]\(u = 4\)[/tex].
When [tex]\(x = 2\)[/tex], [tex]\(u = 8\)[/tex].
The integral now becomes:
[tex]\[ \int_{4}^{8} \frac{1}{\sqrt{u}} \, du = \int_{4}^{8} u^{-\frac{1}{2}} \, du \][/tex]
Evaluate the integral:
[tex]\[ \int u^{-\frac{1}{2}} \, du = 2u^{\frac{1}{2}} \][/tex]
So we have:
[tex]\[ 2u^{\frac{1}{2}} \Big|_4^8 = 2\left(8^{\frac{1}{2}} - 4^{\frac{1}{2}}\right) = 2(\sqrt{8} - \sqrt{4}) = 2(2\sqrt{2} - 2) = 4\sqrt{2} - 4 \][/tex]
4. Solve the second part [tex]\(\int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} \, dx\)[/tex]:
Let [tex]\(x = 2 \tan \theta\)[/tex]. Then, [tex]\(dx = 2 \sec^2 \theta \, d\theta\)[/tex].
When [tex]\(x = 0\)[/tex], [tex]\(\theta = 0\)[/tex].
When [tex]\(x = 2\)[/tex], [tex]\(\theta = \frac{\pi}{4}\)[/tex].
The integral now becomes:
[tex]\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{\sqrt{4 \tan^2 \theta + 4}} \cdot 2 \sec^2 \theta \, d\theta = \int_{0}^{\frac{\pi}{4}} \frac{2 \sec^2 \theta}{2 \sec \theta} \, d\theta = \int_{0}^{\frac{\pi}{4}} \sec \theta \, d\theta \][/tex]
Evaluate the integral:
[tex]\[ \int \sec \theta \, d\theta = \ln|\sec \theta + \tan \theta| \][/tex]
So we have:
[tex]\[ \ln|\sec \theta + \tan \theta| \Big|_0^{\frac{\pi}{4}} = \ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \ln|\sec(0) + \tan(0)| = \ln|\sqrt{2} + 1| - \ln|1| = \ln(\sqrt{2} + 1) - 0 = \ln(\sqrt{2} + 1) \][/tex]
Combining both parts, the total value of the integral is:
[tex]\[ 4\sqrt{2} - 4 + \ln(\sqrt{2} + 1) \approx 2.53822783651192 \][/tex]
So, the evaluated integral [tex]\(\int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx\)[/tex] equals [tex]\(\boxed{2.53822783651192}\)[/tex].
First, identify the integrand: [tex]\(\frac{2x + 1}{\sqrt{x^2 + 4}}\)[/tex].
To solve this integral, we can use techniques such as substitution and recognizing standard forms of integrals. Here is a detailed step-by-step solution:
1. Recognize that the expression under the square root [tex]\(x^2 + 4\)[/tex] suggests using a trigonometric or hyperbolic substitution. However, we can also consider a more straightforward method by splitting the integrand.
2. Split the integrand:
[tex]\[ \int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx = \int_{0}^{2} \frac{2x}{\sqrt{x^2 + 4}} \, dx + \int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} \, dx \][/tex]
3. Solve the first part [tex]\(\int_{0}^{2} \frac{2x}{\sqrt{x^2 + 4}} \, dx\)[/tex]:
Let [tex]\(u = x^2 + 4\)[/tex]. Then, [tex]\(du = 2x \, dx\)[/tex].
When [tex]\(x = 0\)[/tex], [tex]\(u = 4\)[/tex].
When [tex]\(x = 2\)[/tex], [tex]\(u = 8\)[/tex].
The integral now becomes:
[tex]\[ \int_{4}^{8} \frac{1}{\sqrt{u}} \, du = \int_{4}^{8} u^{-\frac{1}{2}} \, du \][/tex]
Evaluate the integral:
[tex]\[ \int u^{-\frac{1}{2}} \, du = 2u^{\frac{1}{2}} \][/tex]
So we have:
[tex]\[ 2u^{\frac{1}{2}} \Big|_4^8 = 2\left(8^{\frac{1}{2}} - 4^{\frac{1}{2}}\right) = 2(\sqrt{8} - \sqrt{4}) = 2(2\sqrt{2} - 2) = 4\sqrt{2} - 4 \][/tex]
4. Solve the second part [tex]\(\int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} \, dx\)[/tex]:
Let [tex]\(x = 2 \tan \theta\)[/tex]. Then, [tex]\(dx = 2 \sec^2 \theta \, d\theta\)[/tex].
When [tex]\(x = 0\)[/tex], [tex]\(\theta = 0\)[/tex].
When [tex]\(x = 2\)[/tex], [tex]\(\theta = \frac{\pi}{4}\)[/tex].
The integral now becomes:
[tex]\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{\sqrt{4 \tan^2 \theta + 4}} \cdot 2 \sec^2 \theta \, d\theta = \int_{0}^{\frac{\pi}{4}} \frac{2 \sec^2 \theta}{2 \sec \theta} \, d\theta = \int_{0}^{\frac{\pi}{4}} \sec \theta \, d\theta \][/tex]
Evaluate the integral:
[tex]\[ \int \sec \theta \, d\theta = \ln|\sec \theta + \tan \theta| \][/tex]
So we have:
[tex]\[ \ln|\sec \theta + \tan \theta| \Big|_0^{\frac{\pi}{4}} = \ln|\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})| - \ln|\sec(0) + \tan(0)| = \ln|\sqrt{2} + 1| - \ln|1| = \ln(\sqrt{2} + 1) - 0 = \ln(\sqrt{2} + 1) \][/tex]
Combining both parts, the total value of the integral is:
[tex]\[ 4\sqrt{2} - 4 + \ln(\sqrt{2} + 1) \approx 2.53822783651192 \][/tex]
So, the evaluated integral [tex]\(\int_{0}^{2} \frac{2x + 1}{\sqrt{x^2 + 4}} \, dx\)[/tex] equals [tex]\(\boxed{2.53822783651192}\)[/tex].
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