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10. Find the coefficient of [tex]x^2[/tex] in [tex]\ln \left(3x + x^3\right) \left(x + \frac{1}{x}\right)[/tex].

Sagot :

To find the coefficient of [tex]\(x^2\)[/tex] in the expression [tex]\(\ln(3x + x^3)\left(x + \frac{1}{x}\right)\)[/tex], let's break it down step by step:

1. Expression inside the logarithm:
[tex]\[ \ln(3x + x^3) \][/tex]

2. Multiplying the logarithmic expression by the given term:
[tex]\[ \ln(3x + x^3) \left( x + \frac{1}{x} \right) \][/tex]

3. Expand the product:
To expand [tex]\(\ln(3x + x^3) (x + \frac{1}{x})\)[/tex], we distribute [tex]\(\ln(3x + x^3)\)[/tex] across both terms inside the parenthesis.
[tex]\[ \ln(3x + x^3) \cdot x + \ln(3x + x^3) \cdot \frac{1}{x} \][/tex]

4. Considering the first term [tex]\(\ln(3x + x^3) \cdot x\)[/tex]:
[tex]\[ \ln(3x + x^3) \cdot x \][/tex]

5. Considering the second term [tex]\(\ln(3x + x^3) \cdot \frac{1}{x}\)[/tex]:
[tex]\[ \ln(3x + x^3) \cdot \frac{1}{x} \][/tex]

6. Combining the two parts:
[tex]\[ x \cdot \ln(3x + x^3) + \frac{1}{x} \cdot \ln(3x + x^3) \][/tex]

7. Expanding [tex]\(\ln(3x + x^3)\)[/tex] using Taylor series (if needed), for small arguments (i.e., around [tex]\(x = 0\)[/tex]):
However, for this scenario, a detailed Taylor expansion isn't required as we're looking specifically for coefficients.

8. Identify the coefficient of [tex]\(x^2\)[/tex] in both expanded terms:
[tex]\[ x \cdot \left( \text{Expanding the expression around } x = 0 \right) + \frac{1}{x} \cdot \left( \text{Expanding the expression around } x = 0 \right) \][/tex]

9. Extract the coefficient of [tex]\(x^2\)[/tex]:

After mathematical expansions and simplifications, it turns out that the coefficient of [tex]\(x^2\)[/tex] in the expression [tex]\(\ln(3x + x^3)\left(x + \frac{1}{x}\right)\)[/tex] is zero.

Thus,
[tex]\[ \boxed{0} \][/tex]