<s>[INST] Evaluating $\int_{\pi/3}^{\pi/2}\frac{1}{\sin\left(x\right)-\cos\left(x\right)+1} \operatorname dx$ $$\int \limits_{\pi/3}^{\pi/2}\dfrac{1}{\sin\left(x\right)-\cos\left(x\right)+1} \operatorname dx$$
If I try solving this integral with universal substitutions: $$\tan\left(\frac{x}{2}\right)=t;\:\sin\left(x\right)=\frac{2t}{1+t^2};\:\cos\left(x\right)=\frac{1-t^2}{1+t^2};\:dx=\frac{2dt}{1+t^2}$$
$$\implies 2\int \limits_{\frac{\sqrt{3}}{3}}^1 \frac{\left(\dfrac{dt}{ 1+t^2}\right)}{ \left(\dfrac{2t-1+t^2+1+t^2}{1+t^2}\right)} $$
I get: $\displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:\dfrac{dt}{t\left(t+1\right)}$
The final solution reached according to my calculations is: $-\ln\left(\frac{\sqrt{3}}{3}\right)-\left(\ln\left(2\right)+\ln\left(\frac{\sqrt{3}}{3}\right)+1\right)$
That is incorrect taking into account the answer given by Symbolab: $\:\ln \left(\frac{1+\sqrt{3}}{2}\right)$
Could the mistake be the approach or something with the calculations?
 [/INST] What you have got
$$\displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:\dfrac{dt}{t\left(t+1\right)}$$
is certainly correct. Now,
\begin{align}\displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:\dfrac{dt}{t\left(t+1\right)} & = \displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:dt\left(\frac{1}{t}-\frac{1}{t+1}\right)\\ & = \log t \biggl|_{\sqrt{3}/3}^1 - \log(1+t)\biggl|_{\sqrt{3}/3}^1 \\ & = \log 1 - \log\frac{\sqrt{3}}{3} -\left(\log 2 - \log\left(1+\frac{\sqrt{3}}{3}\right)\right) \end{align}
and this simplifies to the correct answer.
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