# Compute this integral int_(3^(1/2))^(35^(1/2))(x^2+1)^(1/2)/xdx?

May 25, 2018

$4 + \frac{1}{2} \ln \left(\frac{15}{7}\right)$

#### Explanation:

We begin with the substitution $x = \tan \theta$. Then $\mathrm{dx} = {\sec}^{2} \theta d \theta$ and $\sqrt{{x}^{2} + 1} = \sec \theta$ Thus

${\int}_{{3}^{\frac{1}{2}}}^{{35}^{\frac{1}{2}}} {\left({x}^{2} + 1\right)}^{\frac{1}{2}} / x \mathrm{dx} = {\int}_{{\theta}_{1}}^{{\theta}_{2}} \sec \frac{\theta}{\tan} \theta {\sec}^{2} \theta d \theta$

where ${\theta}_{1} = {\tan}^{-} 1 \sqrt{3}$ and ${\theta}_{2} = {\tan}^{-} 1 \sqrt{35}$

${\int}_{{\theta}_{1}}^{{\theta}_{2}} {\sec}^{3} \frac{\theta}{\tan} \theta d \theta = {\int}_{{\theta}_{1}}^{{\theta}_{2}} \frac{d \theta}{\sin \theta {\cos}^{2} \theta}$
$q \quad q \quad = {\int}_{{\theta}_{1}}^{{\theta}_{2}} \frac{\sin \theta d \theta}{{\sin}^{2} \theta {\cos}^{2} \theta}$

We next substitute $\cos \theta = u$ and use the following

• $\sin \theta d \theta = - \mathrm{du}$
• ${\sin}^{2} \theta = 1 - {u}^{2}$
• $\sec {\theta}_{1} = \sqrt{{\tan}^{2} {\theta}_{1} + 1} = 2 \implies \cos {\theta}_{1} = \frac{1}{2}$
• $\sec {\theta}_{2} = \sqrt{{\tan}^{2} {\theta}_{2} + 1} = 6 \implies \cos {\theta}_{2} = \frac{1}{6}$

to reduce our integral to

${\int}_{\frac{1}{2}}^{\frac{1}{6}} \frac{- \mathrm{du}}{\left(1 - {u}^{2}\right) {u}^{2}} = {\int}_{\frac{1}{6}}^{\frac{1}{2}} \left(\frac{1}{1 - {u}^{2}} + \frac{1}{u} ^ 2\right) \mathrm{du}$
$q \quad = {\left[\frac{1}{2} \ln \left(\frac{1 + u}{1 - u}\right) - \frac{1}{u}\right]}_{\frac{1}{6}}^{\frac{1}{2}}$

$q \quad = \frac{1}{2} \left\{\ln \left(\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}\right) - \ln \left(\frac{1 + \frac{1}{6}}{1 - \frac{1}{6}}\right)\right\} - \frac{1}{\frac{1}{2}} + \frac{1}{\frac{1}{6}}$
$q \quad = \frac{1}{2} \left\{\ln \left(3\right) - \ln \left(\frac{7}{5}\right)\right\} - 2 + 6$
$q \quad = 4 + \frac{1}{2} \ln \left(\frac{15}{7}\right)$