If `(1)/(1^(2))+(1)/(2^(2))+(1)/(3^(2))+(1)/(4^(2))+……+ oo=(pi^(2))/(6)` and `int_(0)^(1)(ln (l+x))/(x)dx=(3pi^(2))/(k)` then k equals :

Answer

To solve the problem, we need to evaluate the integral \( I = \int_0^1 \frac{\ln(1+x)}{x} \, dx \) and relate it to the given series.

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_0^1 \frac{\ln(1+x)}{x} \, dx \]

### Step 2: Use the Taylor series expansion for \(\ln(1+x)\)
The Taylor series expansion for \(\ln(1+x)\) around \(x = 0\) is:
\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
Dividing by \(x\), we have:
\[ \frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \ldots \]

### Step 3: Substitute the series into the integral
Now we substitute this series into the integral:
\[ I = \int_0^1 \left( 1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \ldots \right) \, dx \]

### Step 4: Integrate term by term
We can integrate term by term:
\[ I = \int_0^1 1 \, dx - \frac{1}{2} \int_0^1 x \, dx + \frac{1}{3} \int_0^1 x^2 \, dx - \frac{1}{4} \int_0^1 x^3 \, dx + \ldots \]

Calculating each integral:
- \(\int_0^1 1 \, dx = 1\)
- \(\int_0^1 x \, dx = \frac{1}{2}\)
- \(\int_0^1 x^2 \, dx = \frac{1}{3}\)
- \(\int_0^1 x^3 \, dx = \frac{1}{4}\)

Thus, we have:
\[ I = 1 - \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{3} - \frac{1}{4} \cdot \frac{1}{4} + \ldots \]

### Step 5: Recognize the series
This series can be rewritten as:
\[ I = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \ldots \]
This is the alternating series of \(\frac{1}{n^2}\).

### Step 6: Relate to the known series
We know from the problem statement that:
\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \]
Thus, we can express \(I\) as:
\[ I = \frac{\pi^2}{6} - 2 \left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \ldots \right) \]

### Step 7: Calculate the sum of the series
The series for the even terms can be expressed as:
\[ \sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{24} \]
Thus:
\[ I = \frac{\pi^2}{6} - 2 \cdot \frac{\pi^2}{24} = \frac{\pi^2}{6} - \frac{\pi^2}{12} = \frac{\pi^2}{12} \]

### Step 8: Relate to the given equation
According to the problem, we have:
\[ I = \frac{3\pi^2}{k} \]
Setting the two expressions for \(I\) equal gives:
\[ \frac{\pi^2}{12} = \frac{3\pi^2}{k} \]

### Step 9: Solve for \(k\)
Cancelling \(\pi^2\) from both sides, we get:
\[ \frac{1}{12} = \frac{3}{k} \]
Cross-multiplying gives:
\[ k = 36 \]

### Final Answer
Thus, the value of \(k\) is:
\[ \boxed{36} \]