# g(x)=int_sqrtx^xtan^(-1)tdt Find? 1) g(sqrt3) 2) g'(sqrt3) 3) g"(sqrt3)

May 6, 2018

#### Explanation:

Here,

$g \left(x\right) = {\int}_{\sqrt{x}}^{x} {\tan}^{-} 1 t \mathrm{dt} = {\int}_{\sqrt{x}}^{x} \left(1\right) {\tan}^{-} 1 t \mathrm{dt}$

$\text{Using "color(blue) "Integration by Parts:}$

$I = {\tan}^{-} 1 t \int 1 \mathrm{dt} - \int \left(\frac{d}{\mathrm{dt}} \left({\tan}^{-} 1 t\right) \int 1 \mathrm{dt}\right) \mathrm{dt}$

$I = {\left[{\tan}^{-} 1 t \times t\right]}_{\sqrt{x}}^{x} - {\int}_{\sqrt{x}}^{x} \frac{t}{1 + {t}^{2}} \mathrm{dt}$

$= {\left[t \cdot {\tan}^{-} 1 t\right]}_{\sqrt{x}}^{x} - \frac{1}{2} {\int}_{\sqrt{x}}^{x} \frac{2 t}{1 + {t}^{2}} \mathrm{dt}$

$= x {\tan}^{-} 1 x - \sqrt{x} {\tan}^{-} 1 \sqrt{x} - \frac{1}{2} {\left[\ln | 1 + {t}^{2} |\right]}_{\sqrt{x}}^{x}$

$g \left(x\right) = x {\tan}^{-} 1 x - \sqrt{x} {\tan}^{-} 1 \sqrt{x} - \frac{1}{2} \left[\ln | 1 + {x}^{2} | - \ln | 1 + x |\right]$

$g \left(x\right) = x {\tan}^{-} 1 x - \sqrt{x} {\tan}^{-} 1 \sqrt{x} - \frac{1}{2} \ln | \frac{1 + {x}^{2}}{1 + x} | \ldots \to \left(A\right)$

:.g(sqrt3)=sqrt3xxpi/3-root(4)3tan^-1root(4)3-1/2ln| (1+sqrt3)/(1+root(4)3)|

From $\left(A\right)$,

$g ' \left(x\right) = {\tan}^{-} 1 x - \frac{1}{2 \sqrt{x}} {\tan}^{-} 1 \sqrt{x}$

$g ' \left(\sqrt{3}\right) = \frac{\pi}{3} - \frac{1}{2 \sqrt[4]{3}} {\tan}^{-} 1 \sqrt[4]{3}$

g"$\left(x\right) = \frac{1}{1 + {x}^{2}} - \frac{1}{4 x \left(1 + x\right)} + \frac{1}{4 {x}^{\frac{3}{2}}} {\tan}^{-} 1 \sqrt{x}$

g"$\left(\sqrt{3}\right) = \frac{1}{4} - \frac{1}{4 \sqrt{3} \left(1 + \sqrt{3}\right)} + \frac{1}{4 {\left(3\right)}^{\frac{3}{4}}} {\tan}^{-} 1 \sqrt[4]{3}$