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 sum and the expression \( \frac{3\pi^2}{k} \).

### Step 1: Rewrite the Integral
We can rewrite the integral using the series expansion of \( \ln(1+x) \):
\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
Thus, we have:
\[ I = \int_0^1 \frac{\ln(1+x)}{x} \, dx = \int_0^1 \left(1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \ldots \right) \, dx \]

### Step 2: Integrate Term by Term
Now 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 these integrals:
\[ \int_0^1 x^n \, dx = \frac{1}{n+1} \]
So we have:
\[ I = 1 - \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{3} - \frac{1}{4} \cdot \frac{1}{4} + \ldots \]
This simplifies to:
\[ I = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \ldots \]

### Step 3: Recognize the Series
The series \( 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \ldots \) can be expressed as:
\[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \]
This series converges to \( \frac{\pi^2}{12} \).

### Step 4: Relate to Given Expression
We have found that:
\[ I = \frac{\pi^2}{12} \]
According to the problem, we have:
\[ I = \frac{3\pi^2}{k} \]
Setting these equal gives:
\[ \frac{\pi^2}{12} = \frac{3\pi^2}{k} \]

### Step 5: Solve for \( k \)
To find \( k \), we can cross-multiply:
\[ k \cdot \frac{\pi^2}{12} = 3\pi^2 \]
Dividing both sides by \( \pi^2 \) (assuming \( \pi^2 \neq 0 \)):
\[ \frac{k}{12} = 3 \]
Thus,
\[ k = 36 \]

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