To find the general integral of the given differential equation $y' x = y + x \cos^2(y/x)$, we can use the substitution $z = \frac{y}{x}$.

First, express $y$ in terms of $z$: $y = zx$ $\frac{dy}{dx} = z + x\frac{dz}{dx}$

Substitute these into the differential equation: $x(z + x\frac{dz}{dx}) = zx + x \cos^2(z)$

Simplify this: $xz + x^2 \frac{dz}{dx} = zx + x \cos^2(z)$

$x^2 \frac{dz}{dx} = x \cos^2(z)$

$x \frac{dz}{dx} = \cos^2(z)$

Separate the variables: $\frac{dz}{\cos^2(z)} = \frac{dx}{x}$

The integral of the left side is: $\int \sec^2(z) dz = \int \frac{dx}{x}$

So, $\tan(z) = \ln|x| + C$

Recalling that $z = \frac{y}{x}$, we have: $\tan\left(\frac{y}{x}\right) = \ln|x| + C$

Thus, the correct answer is: $d. \; \tg(y/x) = \ln x + C$

Would you like more details on this solution or have any questions?

Here are five questions you might want to ask next:

1. How do you handle differential equations with trigonometric functions?
2. What is the method for solving exact differential equations?
3. Can you explain the steps for integrating factor methods in differential equations?
4. How do you solve higher-order differential equations?
5. What are some common mistakes to avoid when solving differential equations?

Tip: When dealing with substitutions in differential equations, always check your new variable and its derivatives carefully to avoid algebraic errors.