Answer
$\large\frac{40}{3}$
Work Step by Step
Let \[I=\int_{2}^{6}\frac{y}{\sqrt{y-2}}dy\]
Clearly 2 is point of infinite discontinuity of integrand $\large\frac{y}{\sqrt{y-2}}$
\[\Rightarrow I=\lim_{t\rightarrow 2^+}\int_{t}^{6}\frac{y}{\sqrt{y-2}}dy\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{y}{\sqrt{y-2}}dy\]
\[I_1=\int\frac{(y-2)+2}{\sqrt{y-2}}dy\]
\[I_1=\int\left[\sqrt{y-2}+2(y-2)^{-\frac{1}{2}}\right]dy\]
\[I_1=\frac{2(y-2)^{\frac{3}{2}}}{3}+2(2)(y-2)^{\frac{1}{2}}+C\]
Where $C$ is constant of integration
\[I_1=\frac{2(y-2)^{\frac{3}{2}}}{3}+4\sqrt{y-2}+C\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow 2^+}\left[\frac{2(y-2)^{\frac{3}{2}}}{3}+4\sqrt{y-2}\right]_{t}^{6}\]
\[I=\lim_{t\rightarrow 2^+}\left[\frac{2}{3}(8)+4(2)-\frac{2(t-2)^{\frac{3}{2}}}{3}-4\sqrt{t-2}\right]\]
\[I=\frac{16}{3}+8=\frac{40}{3}\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=\large\frac{40}{3}$.
