Why $\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$? Why does an integral $$\int \frac{dx}{1-x^2}$$ with the limitless (undefined) interval equal to $$\frac 12\ln\frac{1+x}{1-x},$$ yet an integral $$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}$$ with an interval from $\sin\frac{-5π}{12}$ to $\sin\frac{5π}{12}$ has $$\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$$ without one half attached to ln?
$$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\left.\frac12\log\frac{1+x}{1-x}\right|_{\sin\left(-\frac{5\pi}{12}\right)}^{\sin\left(\frac{5\pi}{12}\right)}=\frac12\log\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}-\frac12\log\frac{1+\sin\left(-\frac{5\pi}{12}\right)}{1-\sin\left(-\frac{5\pi}{12}\right)}=$$$${}$$

$$=\frac12\log\left[\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}\cdot\frac{1-\sin\left(-\frac{5\pi}{12}\right)}{1+\sin\left(-\frac{5\pi}{12}\right)}\right]\;(**)$$

But $\;\sin(-x)=-\sin x\;$ , so...

$$(**)=\frac12\log\left(\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}\right)^2=\log\frac{1+\sin\left(\frac{5\pi}{12}\right)}{1-\sin\left(\frac{5\pi}{12}\right)}$$
Since there are already good answers explaining the computation of the definite integral, let me point out what I consider a misunderstanding of the indefinite integral, thus answering the OP question why.

The primitive or indefinite integral is not a function, but a representation of a whole set of functions, all of them when differentiated produce the original function under the integral sign. All these functions differ by a constant, and that is why the computation of the indefinite integral is usually expressed by choosing any of these primitives plus a constant $C$, i.e.:
$$\int \frac{dx}{1-x^2}=\frac 12\ln\frac{1+x}{1-x}+C$$
but you could also write:
$$\int \frac{dx}{1-x^2}=\frac 12\ln \left(2\frac{1+x}{1-x}\right)+C$$
since these two primitives differ by a constant.

Now you can compute a definite integral by Barrow's rule:
$$\int_a^bf(x) dx=F(b)-F(a)$$
where $F'(x)=f(x)$, i.e. $F$ is any primitive or, if you want, the primitive with any value of the constant $C$.

Since you are free to choose the value of $C$, the "aspect" of the primitive can be deceiving with respect to the value of the definite integral, unless you carry out all the detailed computation.