Simplify Before Subbing

Let the integral be $I$ , then:

$\begin{aligned} I & = \int_\frac{\pi}{6}^{\frac{\pi}{3}} \frac{\cot x \sec^2 x}{\cot x +1 }dx \quad \quad \small \color{#3D99F6}{\times \frac{\tan x}{\tan x}} \\ & = \int_\frac{\pi}{6}^{\frac{\pi}{3}} \frac{\sec^2 x}{1 + \tan x }dx \quad \quad \small \color{#3D99F6}{\text{Let } t = \tan x \quad \Rightarrow dt = \sec^2 x \space dx } \\ & = \int_{\frac 1{\sqrt 3}}^{\sqrt{ 3}} \frac{1}{1 + t }dt \\ & = \ln (1+t)\bigg|_\frac{1}{\sqrt{3}}^{\sqrt{3}} = \ln \left(\frac{1+\sqrt{3}}{1+\frac{1}{\sqrt{3}}}\right) = \ln \sqrt{3} = \boxed{\frac{1}{2}\ln 3} \end{aligned}$

I don't like cot or sec, so I rewrite everything:

$\dfrac{\cot(x)\sec^{2}(x)}{\cot(x)+1}=\dfrac{\,\,\,\frac{1}{\sin(x)\cos(x)}\,\,\,}{\frac{\cos(x)}{\sin(x)}+1}=\dfrac{\,\,\,\frac{1}{\cos(x)}\,\,\,}{\cos(x)+\sin(x)}$

$=\dfrac{1}{\cos(x)\Bigl(\cos(x)+\sin(x)\Bigr)}$

To integrate, it's better to have a sum of simplier fractions:

$\dfrac{1}{\cos(x)\Bigl(\cos(x)+\sin(x)\Bigr)}=\dfrac{\sin(x)}{\cos(x)}+\dfrac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}$

That was a good idea as we have now:

$\dfrac{\sin(x)}{\cos(x)}+\dfrac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}=-\dfrac{\cos'(x)}{\cos(x)}+\dfrac{\bigl(\cos(x)+\sin(x)\bigr)'}{\cos(x)+\sin(x)}$

After integrating, we have:

$-\ln\left(\cos(x)\right)+\ln\Bigl(\cos(x)+\sin(x)\Bigr)=\ln\Bigl(\dfrac{\cos(x)+\sin(x)}{\cos(x)}\Bigr)$ $=\ln\Bigl(1+\tan(x)\Bigr)$

Which evaluates to:

$\ln\Bigl(1+\sqrt{3}\Bigr)-\ln\Bigl(1+\frac{1}{\sqrt{3}}\Bigr)=\ln\Bigl(\dfrac{1+\sqrt{3}}{(\sqrt{3}+1)/\sqrt{3}}\Bigr)=\ln\Bigl(\sqrt{3}\Bigr)=\frac{1}{2}\ln\Bigl(3\Bigr)$ .

First, begin by multiplying the integrand by $\frac{\tan(x)}{\tan(x)}$ . This transforms the integrand from $\frac{\cot(x)\sec^2(x)}{\cot(x) + 1} \hspace{1cm} \text{ to } \hspace{1cm} \frac{\sec^2(x)}{1 + \tan(x)}.$

Now we use the appropriate substitution, $u = 1 + \tan(x)$ , thus transforming the integral into $\int \frac{du}{u}$

Integrating and back-substituting gives us $\ln(1 + \tan(x)) \Big|_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \ln\left(1 + \sqrt{3}\right) - \ln\left(1 + \frac{\sqrt{3}}{3}\right) = \frac{1}{2}\ln(3)$