To solve the integral \( \int \frac{dx}{\sin x - \cos x + \sqrt{2}} \), we will use the Weierstrass substitution, which is a common technique for trigonometric integrals. The substitution involves letting:
\(t = \tan\left(\frac{x}{2}\right)\)
Then we have the transformations:
Substituting these into the integral gives:
\(\int \frac{2 \, dt}{1+t^2} \cdot \frac{1}{\frac{2t}{1+t^2} - \frac{1-t^2}{1+t^2} + \sqrt{2}}\)
This simplifies to:
\(\int \frac{2 \, dt}{2t - (1-t^2) + \sqrt{2}(1+t^2)}\)
Simplifying further, we have:
\(\int \frac{2 \, dt}{t^2 + 2\sqrt{2}t + 1}\)
This is a standard integral format. We complete the square in the denominator:
\(t^2 + 2\sqrt{2}t + 1 = (t + \sqrt{2})^2\)
Thus, the integral becomes:
\(\int \frac{2 \, dt}{(t + \sqrt{2})^2}\)
The antiderivative of this expression is:
\(-\frac{1}{t + \sqrt{2}} + C\)
Substituting back for \(t\), we have:
\(t = \tan\left(\frac{x}{2}\right)\)
Therefore, the solution in terms of \(x\) is:
\(-\frac{1}{\tan\left(\frac{x}{2}\right) + \sqrt{2}} + C\)
Which is equivalent to:
\(-\frac{1}{\sqrt{2}}\cot\left(\frac{x}{2} + \frac{\pi}{8}\right) + C\)
Thus, the correct answer is the option:
$-\frac{1}{\sqrt{2}}\cot\left(\frac{x}{2} + \frac{\pi}{8}\right) + C$
The value of : \( \int \frac{x + 1}{x(1 + xe^x)} dx \).