Question

evaluate this!

Answer 1

Sure.

Answer 2

Use Partial Integration.

Answer 3

i need steps please, im really confused

Answer 4

I know with this instruction, you get nothing,
I take the inside of integral to manipulate \[\frac{ 1 }{ x(2x+1) }= \frac{ A }{ x }+\frac{ B }{ 2x+1 }\]
RHS A(2x+1) +Bx =1from LHS
{we can do that because both sides has the same denominator}
---> 2Ax +A +Bx =1
---> x(2A +B) +A =1
---> A =1 and B = -2 (hopefully you know what I am doing, just balance both side with the same terms)
replace to the integral, you have the equivalent one is\[\int\limits_{\frac{ 1 }{ 4 }}^{\frac{ 1 }{ 2 }}\frac{ 1 }{ x }+\frac{ -2 }{ 2x+1 }dx\]
\[=\int\limits_{\frac{ 1 }{ 4 }}^{\frac{ 1 }{ 2 }}\frac{ 1 }{ x }dx-\int\limits_{\frac{ 1 }{ 4 }}^{\frac{ 1 }{ 2 }}\frac{ 2 }{ 2x+1 }dx\]
life is easy from now, right? hope this helps.