Let y = √x, so that y ² = x and 2y dy = dx. Then the integral becomes
[tex]\displaystyle \int_{25}^{81} \frac{\sqrt x}{x-1}\,\mathrm dx = \int_{\sqrt{25}}^{\sqrt{81}} \frac y{y^2-1}(2y\,\mathrm dy) = 2 \int_5^9 \frac{y^2}{y^2-1}\,\mathrm dy[/tex]
Now,
y ² / (y ² - 1) = 1 + 1 / (y ² - 1) = 1 + 1/2 (1/(y - 1) - 1/(y + 1))
so integrating gives us
[tex]\displaystyle 2\int_5^9\frac{y^2}{y^2-1}\,\mathrm dy= \int_5^9\left(2+\frac1{y-1}-\frac1{y+1}\right)\,\mathrm dy \\\\= (2y+\ln|y-1|-\ln|y+1|)\bigg|_5^9 \\\\= \boxed{8+\ln\left(\dfrac65\right)}[/tex]
