Question

How do you integrate `\frac { d x } { x } = ( x e ^ { 3y } + x ) d y`?

Answer 1

`-1/x=1/3e^(3y)+y+C`

Explanation:

Factor:

`dx/x=x(e^(3y)+1)dy`

Separate variables:

`dx/x^2=(e^(3y)+1)dy`

Integrate both sides independently:

`intdx/x^2=int(e^(3y)+1)dy`

On the left, use `intx^ndx=x^(n+1)/(n+1)` where `n!=-1`.

On the right, for `inte^(3y)dy` use the substitution `u=3y` or consider the reverse chain rule, such that `1/3d/(dy)e^(3y)=e^(3y)`.

The constant of integration can be added on either side.

`-1/x=1/3e^(3y)+y+C`

`y` cannot be solved explicitly, so this is an implicit relation.