Series RL current equation
Everything up to step 7 is correct so starting from there:
$${\int_0^t \frac{dt}{L\mathbin{/}R}=\int_0^i\frac{di}{V\mathbin{/}R - i}}$$
Multiplying both sides by \$-1\$
$${\int_0^t \frac{-dt}{L\mathbin{/}R}=\int_0^i\frac{di}{i - V\mathbin{/}R}}$$
You already have the integral of the LHS:
$${\frac{-t}{L\mathbin{/}R}}$$
To solve the right hand side, we can use integration by substitution (Wikipedia):
$${let\  \ u=i-V\mathbin{/}R}$$
$${thus \ du=di}$$
Pluging \$u\$ back into the RHS of equation 7 yields:
$${\int_0^i \frac{du}{u}}=ln|u| \Big|_0^i$$
substituting \$u=i-V\mathbin{/}R\$
$${ln|i-V\mathbin{/}R|\Big|_0^i=ln|i-V\mathbin{/}R|-ln|0-V\mathbin{/}R|}$$
$${\rightarrow \ ln\Big|\frac{i-V\mathbin{/}R}{-V\mathbin{/}R}\Big|}$$
Equating both sides:
$${\frac{-t}{L\mathbin{/}R}=ln\Big| \frac{i-V\mathbin{/}R}{-V\mathbin{/}R}\Big|}$$
Taking both sides as a power of \$e\$:
$${e^{-t\frac{R}{L}}=-\frac{iR}{V}+1}$$
Solving for \$i\$
$${i=\frac{V}{R}\left(1-e^{-t\frac{R}{L}}\right)}$$


  What if we didn't multiply by -1  


So we start again after step 7:
$${\int_0^t \frac{dt}{L\mathbin{/}R}=\int_0^i\frac{di}{V\mathbin{/}R - i}}$$
You already have the integral of the LHS:
$${\frac{t}{L\mathbin{/}R}}$$
To solve the right hand side, we can use integration by substitution (Wikipedia):
$${let\  \ u=V\mathbin{/}R - i}$$
$${thus \ du=-di}$$
Pluging \$u\$ back into the RHS of equation 7 yields:
$${\int_0^i \frac{-du}{u}}=-ln|u| \Big|_0^i$$
substituting \$u=V\mathbin{/}R - i\$
$${-ln|V\mathbin{/}R - i|\Big|_0^i=-\left(ln|V\mathbin{/}R - i|-ln|V\mathbin{/}R - 0|\right)}$$
$${\rightarrow \ -ln\Big|\frac{V\mathbin{/}R - i}{V\mathbin{/}R}\Big|}$$
By law of logarithms:
$${aln(x) = ln(x^a)}$$
Thus:
$${-ln\Big|\frac{V\mathbin{/}R - i}{V\mathbin{/}R}\Big| = ln\Big|\frac{V\mathbin{/}R}{V\mathbin{/}R - i}\Big|}$$
Equating both sides:
$${\frac{t}{L\mathbin{/}R}=ln\Big|\frac{V\mathbin{/}R}{V\mathbin{/}R - i}\Big|}$$
Taking both sides as a power of \$e\$:
$${e^{t\frac{R}{L}}=\frac{V\mathbin{/}R}{V\mathbin{/}R - i}}$$
Solving for \$i\$
$${\frac{V\mathbin{/}R - i}{V\mathbin{/}R} = \frac{1}{e^{t\frac{R}{L}}}}$$
$${1 - i\frac{R}{V} = e^{-t\frac{R}{L}}}$$
Rearranging:
$${i=\frac{V}{R}\left(1-e^{-t\frac{R}{L}}\right)}$$