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[Instruction]
Evaluate the integral of (x + 2)/(x^2 + 9) dx from x = 0 to x = 2

[Ground truth Answer]
To evaluate the integral of (x + 2)/(x^2 + 9) dx from x = 0 to x = 2, we can use integration by substitution and partial fractions. 

First, we rewrite the integrand as a sum of two fractions:

(x + 2) / (x^2 + 9) = A / (x + 3i) + B / (x - 3i),

where A and B are constants, and i is the imaginary unit. To find A and B, we clear the denominators:

(x + 2) = A(x - 3i) + B(x + 3i).

Setting x = 3i, we get A = (1/2)(3i + 2), and setting x = -3i, we find B = (1/2)(-3i + 2). Thus, our integrand becomes:

(x + 2) / (x^2 + 9) = (1/2)(3i + 2) / (x + 3i) + (1/2)(-3i + 2) / (x - 3i).

Now we can split the integral into two parts and evaluate them separately:

∫[(x + 2) / (x^2 + 9)] dx = (1/2)∫[(3i + 2) / (x + 3i)] dx + (1/2)∫[(-3i + 2) / (x - 3i)] dx.

Let u = x + 3i and v = x - 3i, so du = dv = dx. Then our integral becomes:

(1/2)∫[(3i + 2) / u] du + (1/2)∫[(-3i + 2) / v] dv.

Integrating these gives:

(1/2)[(3i + 2) ln|u|] + (1/2)[(-3i + 2) ln|v|] + C,

where C is the constant of integration. Substituting back for u and v in terms of x, we have:

(1/2)[(3i + 2) ln|x + 3i|] + (1/2)[(-3i + 2) ln|x - 3i|] + C.

Finally, we evaluate this expression at the limits x = 0 and x = 2:

F(2) - F(0) = (1/2)[(3i + 2) ln|5i| - (-3i + 2) ln(3i)] - (1/2)[(3i + 2) ln|3i| - (-3i + 2) ln|-3i|].

The imaginary parts cancel out, leaving:

F(2) - F(0) = (1/2)[2 ln|(5i)(-3i)/(3i)^2]| = ln|5/3|.

So, the value of the integral is ln|5/3|.

[Assistant's Response]
To evaluate the integral of \(\frac{x + 2}{x^2 + 9}\) from \(x = 0\) to \(x = 2\), we can break it down into two separate integrals:

\[
\int_0^2 \frac{x + 2}{x^2 + 9} \, dx = \int_0^2 \frac{x}{x^2 + 9} \, dx + \int_0^2 \frac{2}{x^2 + 9} \, dx
\]

### First Integral: \(\int \frac{x}{x^2 + 9} \, dx\)

For this integral, we can use the substitution \(u = x^2 + 9\), which gives \(du = 2x \, dx\) or \(\frac{1}{2} du = x \, dx\). The limits of integration change as follows:
- When \(x = 0\), \(u = 0^2 + 9 = 9\).
- When \(x = 2\), \(u = 2^2 + 9 = 13\).

Thus, the integral becomes:

\[
\int_9^{13} \frac{1}{2} \cdot \frac{1}{u} \, du = \frac{1}{2} \left[ \ln |u| \right]_9^{13} = \frac{1}{2} \left( \ln 13 - \ln 9 \right) = \frac{1}{2} \ln \frac{13}{9}
\]

### Second Integral: \(\int \frac{2}{x^2 + 9} \, dx\)

This integral can be recognized as a standard form that can be integrated using the arctangent function. The integral of \(\frac{1}{a^2 + x^2}\) is \(\frac{1}{a} \arctan\left(\frac{x}{a}\right) + C\). Here, \(a = 3\), so we have:

\[
\int \frac{2}{x^2 + 9} \, dx = 2 \cdot \frac{1}{3} \int \frac{1}{1 + \left(\frac{x}{3}\right

[The End of Assistant's Response]

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[System]



Assistant: