# How will you integrate ? int(dx)/(1+x^4)^(1/4)

Apr 17, 2017

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

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

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

Let $\left(1 + \frac{1}{x} ^ 4\right) = {u}^{4}$

Differentiating we get

$- \frac{4}{x} ^ 5 \mathrm{dx} = 4 {u}^{3} \mathrm{du}$

$\implies - \frac{\cancel{4}}{x} ^ 5 \mathrm{dx} = \cancel{4} {u}^{3} \mathrm{du}$

So Intgral

$I = - \int \frac{{u}^{3} \mathrm{du}}{\left({u}^{4} - 1\right) u}$

$= - \int \frac{{u}^{2} \mathrm{du}}{\left({u}^{2} - 1\right) \left({u}^{2} + 1\right)}$

$= - \frac{1}{2} \int \frac{\left(\left({u}^{2} + 1\right) + \left({u}^{2} - 1\right)\right) \mathrm{du}}{\left({u}^{2} - 1\right) \left({u}^{2} + 1\right)}$

$= - \frac{1}{2} \int \frac{\mathrm{du}}{{u}^{2} - 1} - \frac{1}{2} \int \frac{\mathrm{du}}{{u}^{2} + 1}$

$= - \frac{1}{4} \int \frac{\left(\left(u + 1\right) - \left(u - 1\right)\right) \mathrm{du}}{{u}^{2} - 1} - \frac{1}{2} \int \frac{\mathrm{du}}{{u}^{2} + 1}$

$= - \frac{1}{4} \int \frac{\mathrm{du}}{u - 1} + \frac{1}{4} \int \frac{\mathrm{du}}{u + 1} - \frac{1}{2} \frac{\mathrm{du}}{{u}^{2} + 1}$

$= - \frac{1}{4} \ln \left\mid u - 1 \right\mid + \frac{1}{4} \ln \left\mid u + 1 \right\mid - \frac{1}{2} {\tan}^{-} 1 u + c$

Inserting $u = {\left({x}^{4} + 1\right)}^{\frac{1}{4}} / x$

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