To integrate `int(2dI)/(10-3I)=intdt` we start by integrating the right side which is `t+C` and we have:

`int(2dI)/(10-3I)=t+C`

Ingetrating `intdt` means take the integral of `int(1)dt` which mean take the integral of `1` and as the variable is `t` in `dt` it refers to taking the drevative of `1` with respect to `t`.

`:.` the `int(1)dt` is `(1)t+C` which can be written as `t+C`

On the left side we have:

`int(2dI)/(10-3I)`

To the integral of `1/x` is `ln|x|` and if we have `1/(2x+3)`, its integral is `ln|2x+3|*1/2`

but if we have `2/x` the integral of `int2/xdx` is `2ln|x|` and pay close attention to 2 that appears as a coefficient of ln(x).

In our case we have `int(2dI)/(10-3I)` so :

`[2*ln|10-3I|*1/3]` which can be written as `[2/3*ln|10-3I|]` (I moved `1/3` to the front).

Now our complete equation looks like this:

`2/3*ln|10-3I|=t+C`

The integration is completed but we want to solve for I.

Solving for IMultiply both side by `3/2` so we get rid of `2/3` from the left side.

`3/2*2/3*ln|10-3I|=(t+C)*3/2` and we have:

`ln|10-3I|=(t+C)*3/2`

Distribute (multiply) the `3/2` on the right side and we have:

`ln|10-3I|=3/2t+3/2C`

Because C is an unknown constant, we write it as `ln|10-3I|=3/2t+C`

Get rid if `ln` from `ln|10-3I|=3/2t+C` using `e` ad the base for both side:

`e^(ln|10-3I|)=e^(3/2t+C)`

`e` cancels the `ln` and we have:

`|10-3I|=e^(3/2t+C)`

on the right side we have `e^(3/2t+C)`. Recall that if we have `A^rxxA^v`, it can be written as `A^(r+v)`. We we have `e^(3/2t+C)` so we can go reverse and write it as `e^(3/2t)*e^C`

We have now:

`|10-3I|=e^(3/2t)*e^C`

to get rid of the absolute value `|10-3I|` we must add `+-` on the other side of equation and we get:

`10-3I=+-e^(3/2t)*e^C`

On the right side we have `+-e^(3/2t)*e^C`. We can rearrange it as `e^(3/2t)*(+-e^C)`. This `(+-e^C)` is still a arbitrary constant so we can replace it for example by `k`

Our equation becomes `10-3I=ke^(3/2t)`