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Ignore question number 10.(i) . Help me answer the number 10(ii) pleaseeeee.

Ignore question number 10.(i) . Help me answer the number 10(ii) pleaseeeee.-example-1
User Jacky Lau
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1 Answer

6 votes

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)

After differentiating in i, we had;


(dy)/(dx)=10\ln(5x+1)+10


We now integrate both sides to obtain;


\int\limits {(dy)/(dx)} \, dx =\int\limits {10\ln(5x+1)+10} \, dx


This gives,



y =\int\limits {10\ln(5x+1)} \, dx+\int\limits {10} \, dx


We now split the integral to obtain;



y =\int\limits {10\ln(5x+1)} \, dx+\int\limits {10} \, dx



This implies that,


y =\int\limits {10\ln(5x+1)} \, dx+10x +C


We rearrange to get,


\int\limits {10\ln(5x+1)} \, dx = y-10x +C


But


y=(10x+2)\ln(5x+1)


This implies,


10
\int\limits {\ln(5x+1)} \, dx = (10x+2)\ln(5x+1)-10x +C


We divide through by 10.



\int\limits {\ln(5x+1)} \, dx = 2((5x+1)\ln(5x+1))/(10)-x +(C)/(10)

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


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










User John Lim
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