Step 1: Plan:
The antiderivative is $I(k)=k-\dfrac{\ln\left(\frac{1+4^k}{2}\right)}{\ln4}$. We test the given options against the target value.
Step 2: Test k = 3/2:
$I(k)=k-\dfrac{\ln\left(\frac{1+4^k}{2}\right)}{\ln4}$. For $k=\tfrac32$, $4^k=8$, so $\dfrac{1+8}{2}=4.5$ and $\ln4.5=1.504$. $\dfrac{1.504}{1.386}=1.085$. Then $I=1.5-1.085=0.415$.
Step 3: Compare with the target:
$\log_2\dfrac43=\dfrac{\ln1.3333}{\ln2}=\dfrac{0.2877}{0.6931}=0.415$. They match, so $k=\tfrac32$, option B.
Final Answer:
Integrating gives k - log base 4 of (1 + 4^k)/2, which is solved by k = 3/2.
\[ \boxed{\text{(B) }\dfrac32} \]