$\text{The value of the integral}{\int }_0^{\frac{\pi }{4}}{log}_e(1+tanx)d\mathrm{<mtext>x\; is:</mtext>\; }$

The correct answer is:
$$\frac{\pi }{16}{log}_{e}4$$

Step-by-step Derivation and Explanation:

Let the given definite integral be denoted by I:
$$I={\int }_0^{\frac{\pi }{4}}{log}_e(1+tanx)dx$$ --- (Equation 1)

To solve this, we can apply the properties of definite integrals. Specifically, we use the property:
$${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$$

Here, the lower limit a is 0 and the upper limit b is $\frac{\pi }{4}$. Therefore, we substitute x with $\frac{\pi }{4}-x$:
$$I={\int }_0^{\frac{\pi }{4}}{log}_e\left[1+tan\left(\frac{\pi }{4}-x\right)\right]dx$$

Using the trigonometric identity for the tangent of a difference:
$$tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}$$

Since $tan\left(\frac{\pi }{4}\right)=1$, we have:
$$tan\left(\frac{\pi }{4},-,x\right)=\frac{1-tanx}{1+tanx}$$

Substitute this back into the expression for I:
$$I={\int }_0^{\frac{\pi }{4}}{log}_e\left[1+\frac{1-tanx}{1+tanx}\right]dx$$

Simplify the term inside the logarithm:
$$1+\frac{1-tanx}{1+tanx}=\frac{(1+tanx)+(1-tanx)}{1+tanx}=\frac{2}{1+tanx}$$

Now substitute this back into the integral:
$$I={\int }_0^{\frac{\pi }{4}}{log}_e\left(\frac{2}{1+tanx}\right)dx$$ --- (Equation 2)

Using logarithmic properties (${log}_e\left(\frac{A}{B}\right)={log}_{e}A-{log}_{e}B$), we get:
$$I={\int }_0^{\frac{\pi }{4}}\left[{log}_{e}2-{log}_e(1+tanx)\right]dx$$

Splitting the integral:
$$I={\int }_0^{\frac{\pi }{4}}{log}_{e}2dx-{\int }_0^{\frac{\pi }{4}}{log}_e(1+tanx)dx$$

Notice that the second integral is our original integral I:
$$I={log}_{e}2{\int }_0^{\frac{\pi }{4}}1dx-I$$

Add I to both sides:
$$2I={log}_{e}2{\left[x\right]}_0^{\frac{\pi }{4}}$$

Evaluate the limits:
$$2I={log}_{e}2\left(\frac{\pi }{4}-0\right)$$
$$2I=\frac{\pi }{4}{log}_{e}2$$

Divide by 2 to find I:
$$I=\frac{\pi }{8}{log}_{e}2$$

To write the answer in the form of the given options, we can rewrite ${log}_{e}2$ using the property $k{log}_{e}x={log}_e\left(x^k\right)$:
$$I=\frac{\pi }{16}\times 2{log}_{e}2$$
$$I=\frac{\pi }{16}{log}_e\left(2^2\right)$$
$$I=\frac{\pi }{16}{log}_{e}4$$