If _0^40 dx/(2x+1) = _e k, then the value of k is : Option 1: 3 Option 2: 9/2 Option 3: 9 Option 4: 3/2

Answer: 9. Solution: _0^40dx/(2x+1)=12[_e(2x+1)]_0^40=12_e 81=_e 9. Hence k=9.

Calculus

Integrals

Easy

common

If $\displaystyle\int_0^{40} \dfrac{dx}{2x+1} = \log_e k$ , then the value of $k$ is :

$\dfrac{9}{2}$

$\dfrac{3}{2}$

✅ Correct Option: 3

$\displaystyle\int_0^{40}\dfrac{dx}{2x+1}=\dfrac12\left[\log_e(2x+1)\right]_0^{40}=\dfrac12\log_e 81=\log_e 9$ . Hence $k=9$ .

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