EXPLANATION
The ii says "hence". This tells you that you must proceed from i above.
Given;
[tex]y=(10x+2)\ln(5x+1)[/tex]
After differentiating in i, we had;
[tex]\frac{dy}{dx}=10\ln(5x+1)+10[/tex]
We now integrate both sides to obtain;
[tex]\int\limits {\frac{dy}{dx}} \, dx =\int\limits {10\ln(5x+1)+10} \, dx[/tex]
This gives,
[tex]y =\int\limits {10\ln(5x+1)} \, dx+\int\limits {10} \, dx[/tex]
We now split the integral to obtain;
[tex]y =\int\limits {10\ln(5x+1)} \, dx+\int\limits {10} \, dx [/tex]
This implies that,
[tex]y =\int\limits {10\ln(5x+1)} \, dx+10x +C[/tex]
We rearrange to get,
[tex]\int\limits {10\ln(5x+1)} \, dx = y-10x +C[/tex]
But
[tex]y=(10x+2)\ln(5x+1)[/tex]
This implies,
10[tex]\int\limits {\ln(5x+1)} \, dx = (10x+2)\ln(5x+1)-10x +C[/tex]
We divide through by 10.
[tex]\int\limits {\ln(5x+1)} \, dx = 2\frac{(5x+1)\ln(5x+1)}{10}-x +\frac{C}{10}[/tex]
NB: The constant C divided by 10 is still a constant.
[tex]\int\limits {\ln(5x+1)} \, dx = \frac{(5x+1)\ln(5x+1)}{5}-x +C[/tex]
If a=5 and b=1 and we substitute, we get a general formula,but they were partially substituted to get.
[tex]\int\limits {\ln(5x+1)} \, dx = \frac{(ax+b)\ln(5x+1)}{5}-x +C[/tex]
