# How do you find the integral of dx/((x^(5/2) * (1+ x^2)^(1/4))?

Jul 21, 2018

$\int \frac{1}{{x}^{\frac{5}{2}} {\left(1 + {x}^{2}\right)}^{\frac{1}{4}}} \mathrm{dx} = - \frac{2}{3} {\left(\frac{1}{x} ^ 2 + 1\right)}^{\frac{3}{4}} + c$

#### Explanation:

Here ,

$I = \int \frac{1}{{x}^{\frac{5}{2}} {\left(1 + {x}^{2}\right)}^{\frac{1}{4}}} \mathrm{dx}$

Substitute color(red)(x=1/t=>dx=-1/t^2dt

So,

$I = \int \frac{1}{{\left(\frac{1}{t}\right)}^{\frac{5}{2}} {\left(1 + \frac{1}{t} ^ 2\right)}^{\frac{1}{4}}} \left(- \frac{1}{t} ^ 2\right) \mathrm{dt}$

$= - \int {t}^{\frac{5}{2}} / \left({\left({t}^{2} + 1\right)}^{\frac{1}{4}} / {t}^{\frac{1}{2}}\right) \cdot \frac{1}{t} ^ 2 \mathrm{dt}$

$= - \int {t}^{\frac{5}{2} + \frac{1}{2} - 2} / {\left({t}^{2} + 1\right)}^{\frac{1}{4}} \mathrm{dt}$

$\therefore I = - \int \frac{t}{{t}^{2} + 1} ^ \left(\frac{1}{4}\right) \mathrm{dt}$

Substitute color(blue)((t^2+1)^(1/4)=u=>t^2+1=u^4

$\implies 2 t \mathrm{dt} = 4 {u}^{3} \mathrm{du} \implies t \mathrm{dt} = 2 {u}^{3} \mathrm{du}$

$I = - \int \frac{2 {u}^{3}}{u} \mathrm{du}$

$\implies I = - 2 \int {u}^{2} \mathrm{du}$

$\implies I = - 2 \left[{u}^{3} / 3\right] + c$

$= - \frac{2}{3} {\left(u\right)}^{3} + c$

Subst. back , color(blue)(u=(t^2+1)^(1/4)

$I = - \frac{2}{3} {\left({t}^{2} + 1\right)}^{\frac{3}{4}} + c$

Again subst. color(red)(t=1/x

$\therefore I = - \frac{2}{3} {\left(\frac{1}{x} ^ 2 + 1\right)}^{\frac{3}{4}} + c$

$\implies I = - \frac{2}{3} {\left(1 + {x}^{2}\right)}^{\frac{3}{4}} / {\left({x}^{2}\right)}^{\frac{3}{4}} + c$

$\therefore I = - \frac{2 {\left(1 + {x}^{2}\right)}^{\frac{3}{4}}}{3 {x}^{\frac{3}{2}}} + c$