If ∫04dx2x+1=log⁡k\int_0^4 \frac{dx}{2x + 1} = \log k, then the value of k is :

Answer: 3

Step-by-step solution

Step 1: Integrate the function

We need to evaluate the definite integral. The integral of 1/(ax+b)1/(ax+b) is (1/a)log⁡∣ax+b∣(1/a) \log |ax+b|. In this case, a=2a=2 and b=1b=1.

Step 2: Apply the limits of integration

Now we apply the limits of integration from 00 to 44. This means we substitute the upper limit into the integrated function and subtract the result of substituting the lower limit.

Step 3: Evaluate the definite integral

Substitute x=4x=4 and x=0x=0 into the integrated expression. This gives us 12log⁡(9)−12log⁡(1)\frac{1}{2} \log(9) - \frac{1}{2} \log(1).

Step 4: Simplify the expression

Since log⁡1=0\log 1 = 0, the expression simplifies to 12log⁡9\frac{1}{2} \log 9. We can also write 12log⁡9\frac{1}{2} \log 9 as log⁡(91/2)\log (9^{1/2}) which is log⁡3\log 3.

Step 5: Find the value of k

We are given that the integral equals log⁡k\log k. By comparing our result log⁡3\log 3 with log⁡k\log k, we find that k=3k=3.

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