We start with:

`Ri+L(di)/(dt)=V`

Subtracting Ri from both sides:

`L(di)/(dt)=V-Ri`

Divide both sides by L:

`(di)/(dt)=(V-Ri)/L`

Multiply both sides by dt and divide both by (V - Ri):

`(di)/(V-Ri)=(dt)/L`

Integrate (see Integration: Basic Logarithm Form):

`int(di)/(V-Ri)=int(dt)/L`

`-(ln(V-Ri))/R=1/Lt+K`

Now, since `i = 0` when `t = 0`, we have:

`K=-(ln\ V)/R`

Substituting K back into our expression:

`-(ln(V-Ri))/R=1/Lt-(ln V)/R`

Rearranging:

`(ln\ V)/R-(ln(V-Ri))/R=1/Lt`

Multiplying throughout by -R:

`-ln\ V+ln(V-Ri)=-R/Lt`

Collecting the logarithm parts together:

`ln((V-Ri)/V)=-R/Lt`

Taking "e to both sides":

`(V-Ri)/V=e^(-(R"/"L)t`

`1-R/Vi=e^(-(R"/"L)t`

Subtracting 1 from both sides:

`-R/Vi=-1+e^(-(R"/"L)t`

Multiplying both sides by `-(V/R)`:

`i=V/R(1-e^(-(R"/"L)t))`