Question

integral dy= integral ((y+square root(( y^2 - 8x)/4x))dx

Answer 1

To solve the integral

\[
\int dy = \int \left( y + \sqrt{\frac{y^2 - 8x}{4x}} \right) dx,
\]

we will first separate the variables.

From the left side, we simply integrate with respect to \(y\):

\[
\int dy = y + C_1,
\]

where \(C_1\) is an arbitrary constant of integration.

Now, let's focus on the right side:

\[
\int \left( y + \sqrt{\frac{y^2 - 8x}{4x}} \right) dx.
\]

We can split this into two separate integrals:

\[
\int y \, dx + \int \sqrt{\frac{y^2 - 8x}{4x}} \, dx.
\]

The term \(y\) is treated as a constant when integrating with respect to \(x\). So

\[
\int y \, dx = yx + C_2,
\]

where \(C_2\) is another arbitrary constant of integration.

Now we can evaluate the second integral. To simplify \(\sqrt{\frac{y^2 - 8x}{4x}} = \frac{\sqrt{y^2 - 8x}}{2\sqrt{x}}\). Thus,

\[
\int \sqrt{\frac{y^2 - 8x}{4x}} \, dx = \int \frac{\sqrt{y^2 - 8x}}{2\sqrt{x}} \, dx.
\]

The integral \(\int \sqrt{y^2 - 8x} \, dx\) can be more complicated and typically involves substitution or integration by parts, depending on the specific context or the function of \(y\).

To summarize, we have:

\[
y + C_1 = yx + \int \frac{\sqrt{y^2 - 8x}}{2\sqrt{x}} \, dx + C_2.
\]

At this point, we can combine the constant terms \(C_1 - C_2\) and rewrite the equation:

\[
y = yx + \int \frac{\sqrt{y^2 - 8x}}{2\sqrt{x}} \, dx + C.
\]

Where \(C = C_1 - C_2\).

This gives us an implicit relationship between \(x\) and \(y\). Depending on your specific requirements (whether you seek \(y\) in explicit form, or wish to evaluate for a specific \(x\)), further steps may be needed.