Step-by-step explanation
The ii says "hence". This tells you that you must proceed from i above.
Given;
![y=(10x+2)\ln(5x+1)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/sp24xmj0io6hp4d3islurk9wcsotduvwa7.png)
After differentiating in i, we had;
![(dy)/(dx)=10\ln(5x+1)+10](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ihpjmbid1cf2jjla8nqbuzfgticzq4at6b.png)
We now integrate both sides to obtain;
![\int\limits {(dy)/(dx)} \, dx =\int\limits {10\ln(5x+1)+10} \, dx](https://img.qammunity.org/2019/formulas/mathematics/middle-school/d6aiazf1vn49a45na3pps01ss3v9gi923h.png)
This gives,
![y =\int\limits {10\ln(5x+1)} \, dx+\int\limits {10} \, dx](https://img.qammunity.org/2019/formulas/mathematics/middle-school/jaresukuwv320yrly1wr5c4m1cmiuwrq5j.png)
We now split the integral to obtain;
![y =\int\limits {10\ln(5x+1)} \, dx+\int\limits {10} \, dx](https://img.qammunity.org/2019/formulas/mathematics/middle-school/jaresukuwv320yrly1wr5c4m1cmiuwrq5j.png)
This implies that,
![y =\int\limits {10\ln(5x+1)} \, dx+10x +C](https://img.qammunity.org/2019/formulas/mathematics/middle-school/3lddsy8gtlr6hnysvipoacde0o3ajr1v4i.png)
We rearrange to get,
![\int\limits {10\ln(5x+1)} \, dx = y-10x +C](https://img.qammunity.org/2019/formulas/mathematics/middle-school/7umyxkvx1khcpyhwjiuaxr5vrnb7ud6knd.png)
But
![y=(10x+2)\ln(5x+1)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/sp24xmj0io6hp4d3islurk9wcsotduvwa7.png)
This implies,
10
![\int\limits {\ln(5x+1)} \, dx = (10x+2)\ln(5x+1)-10x +C](https://img.qammunity.org/2019/formulas/mathematics/middle-school/odbd1qlgdel32iibrg57j7mfjmdxdxf2kz.png)
We divide through by 10.
![\int\limits {\ln(5x+1)} \, dx = 2((5x+1)\ln(5x+1))/(10)-x +(C)/(10)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/22hqocknu87tahhk37o9btqs0czagv5db2.png)
NB: The constant C divided by 10 is still a constant.
![\int\limits {\ln(5x+1)} \, dx = ((5x+1)\ln(5x+1))/(5)-x +C](https://img.qammunity.org/2019/formulas/mathematics/middle-school/4ejnsrr3gjtxtqg0oaylo2h6uczzggoktp.png)
If a=5 and b=1 and we substitute, we get a general formula,but they were partially substituted to get.
![\int\limits {\ln(5x+1)} \, dx = ((ax+b)\ln(5x+1))/(5)-x +C](https://img.qammunity.org/2019/formulas/mathematics/middle-school/xe5fuhxdfp5c3r6nwrgbk1ftq9skmlbrb7.png)