`int _(0) ^(pi//4) (xdx )/(cos x (cos x + sin x ))`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{x}{\cos x (\cos x + \sin x)} \, dx, \]

we will use a property of definite integrals. The property states that

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

Here, we have \( a = 0 \) and \( b = \frac{\pi}{4} \). Therefore, we will replace \( x \) with \( \frac{\pi}{4} - x \):

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\frac{\pi}{4} - x}{\cos\left(\frac{\pi}{4} - x\right) \left(\cos\left(\frac{\pi}{4} - x\right) + \sin\left(\frac{\pi}{4} - x\right)\right)} \, dx. \]

Next, we need to evaluate \( \cos\left(\frac{\pi}{4} - x\right) \) and \( \sin\left(\frac{\pi}{4} - x\right) \):

Using the angle subtraction formulas:

\[ \cos\left(\frac{\pi}{4} - x\right) = \cos\frac{\pi}{4}\cos x + \sin\frac{\pi}{4}\sin x = \frac{1}{\sqrt{2}}(\cos x + \sin x), \]

\[ \sin\left(\frac{\pi}{4} - x\right) = \sin\frac{\pi}{4}\cos x - \cos\frac{\pi}{4}\sin x = \frac{1}{\sqrt{2}}(\cos x - \sin x). \]

Now substituting these into our expression for \( I \):

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\frac{\pi}{4} - x}{\frac{1}{\sqrt{2}}(\cos x + \sin x) \left(\frac{1}{\sqrt{2}}(\cos x + \sin x)\right)} \, dx. \]

This simplifies to:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\frac{\pi}{4} - x}{\frac{1}{2}(\cos x + \sin x)^2} \, dx = 2 \int_{0}^{\frac{\pi}{4}} \frac{\frac{\pi}{4} - x}{(\cos x + \sin x)^2} \, dx. \]

Now we can add the two expressions for \( I \):

\[ 2I = \int_{0}^{\frac{\pi}{4}} \frac{x + \left(\frac{\pi}{4} - x\right)}{\cos x (\cos x + \sin x)} \, dx = \int_{0}^{\frac{\pi}{4}} \frac{\frac{\pi}{4}}{\cos x (\cos x + \sin x)} \, dx. \]

Thus,

\[ I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos x (\cos x + \sin x)} \, dx. \]

Next, we simplify the integral:

\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos x (\cos x + \sin x)} \, dx. \]

We can factor out \( \cos x \):

\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos^2 x} \cdot \frac{1}{1 + \tan x} \, dx = \int_{0}^{\frac{\pi}{4}} \sec^2 x \cdot \frac{1}{1 + \tan x} \, dx. \]

Now we make the substitution \( t = \tan x \), which gives \( dt = \sec^2 x \, dx \). The limits change as follows: when \( x = 0 \), \( t = 0 \) and when \( x = \frac{\pi}{4} \), \( t = 1 \):

\[ \int_{0}^{1} \frac{1}{1 + t} \, dt = \left[ \ln(1 + t) \right]_{0}^{1} = \ln(2) - \ln(1) = \ln(2). \]

Thus, we have:

\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos x (\cos x + \sin x)} \, dx = \ln(2). \]

Finally, substituting back, we find:

\[ I = \frac{\pi}{8} \ln(2). \]

### Final Answer:
\[ I = \frac{\pi}{8} \ln(2). \]