Question

integral of [7y+2]/[3+4y+4y^2]dy...need help pls

Answer 1

\[\int\limits \frac{ 7y+2 }{ 3+4y+4y^2 } dy \]

This one see the denominator is a degree less than the numerator.
So we expect to possibly get a log out of it and perfect square --> inverse tangent

Let's differentiate the denominator to get 8y + 4. So let's make this 8y + 4 appear on the numerator. Watch the steps carefully to balance...

\[\int\limits \frac{ 7y+2 }{ 3+4y+4y^2 } dy = \frac{ 7 }{ 8 } \int\limits \frac{ (8y+16/7) }{ 3+4y+4y^2}dy= \frac{ 7 }{ 8 } \int\limits \frac{ 8y+4 + \frac{ 1 }{ 7 } }{ 3+4y+4y^2 }dy\]

Then we see the denominator can be made into a perfect square
\[4y^2+4y+3 = 4y^2+4y + 1 + 2 = (2y+1)^2+3\]

Recall
\[\int\limits \frac{ f'(x) }{ a^2+(f(x))^2 } = \frac{ 1 }{ a } \tan^{-1} \frac{ f(x) }{ a } +C\]
Then combine the log and arctan results.