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

Question

asked Nov 1, 2019 226k views

1 Answer

John Lim · Answer 1 · 2019-11-06T02:35:40+0000

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.

$\int\limits {\ln(5x+1)} \, dx = ((ax+b)\ln(5x+1))/(5)-x +C$