Help with $\int \frac{1}{(\sin x + \cos x)}$ Kindly solve this question
$$\int \frac{1}{(\sin x + \cos x)} dx$$
I reached up to 
$$\frac{(1+\tan^2x)}{1-\tan^2x + 2\tan x}$$

Hint :
\begin{align}
\int \frac{\mathrm dx}{\sin x + \cos x} &=\int \frac{\mathrm dx}{\sin x + \cos x}\cdot \frac{\sin x - \cos x}{\sin x - \cos x}\\
&=\int \frac{\sin x - \cos x}{\sin^2 x - \cos^2 x}\mathrm dx\\
&=\int \frac{\sin x}{1 - 2\cos^2 x}\mathrm dx-\int \frac{\cos x}{2\sin^2 x -1}\mathrm dx\\
&=\int \frac{1}{2\cos^2 x-1}\mathrm d(\cos x)-\int \frac{1}{2\sin^2 x -1}\mathrm d(\sin x)\\
\end{align}
