Question

medal + fan
please asap

Answer 1

\[\int\limits_{}^{} \frac{ y }{ y^2 + y + 1 } dy\]

Answer 2

@ganeshie8

Answer 3

\[\displaystyle I = \int \frac{y}{y^2+y+1}dy = \frac{1}{2}\int\frac{(2y+1) -1}{y^2+y+1}dy\]
\[\displaystyle 2I = \int \frac{2y+1}{y^2+y+1}dy - \int\frac{dy}{y^2+y+1}\]
Now, if \(t = y^2 + y +1 \implies dt = (2y+1)dy\) So, the first integral reduces to the form \(\displaystyle \int \frac{dt}{t }= ln(|t|) \)

For the second integral, note that \(\displaystyle y^2 + y +1 = \left( y + \frac{1}{2} \right )^2 + \frac{3}{4}\), so let \( \displaystyle y + \frac{1}{2} = \frac{\sqrt{3}}{2} \tan \theta \implies dy = \frac{\sqrt{3}}{2}\sec ^2{\theta} d\theta\)
\[\int \frac{dy}{y^2+y+1} = \frac{(\sqrt{3}/2)\sec^2\theta}{(3/4)(1+\tan^2 \theta)} d\theta = \frac{2}{\sqrt{3}}\int d\theta = \frac{2}{\sqrt{3}} \theta \]

I hope this is enough :)