# Find the integral ?

Jun 27, 2018

$\frac{1}{4} \arcsin h \left({x}^{4} / 2\right) + C$

#### Explanation:

int (x^3*dx)/(sqrt(x^8+4)

=1/4int (4x^3*dx)/(sqrt((x^4)^2+4)

After using ${x}^{4} = 2 \sinh u$ and $4 {x}^{4} \cdot \mathrm{dx} = 2 \cosh u \cdot \mathrm{du}$ transforms,
this integral became

$\frac{1}{4} \int \frac{2 \cosh u \cdot \mathrm{du}}{2 \cosh u}$

=$\frac{1}{4} \int \mathrm{du}$

=1/4*u+C#

After using ${x}^{4} = 2 \sinh u$ and $u = \arcsin h \left({x}^{4} / 2\right)$ inverse transforms, I found

$\frac{1}{4} \arcsin h \left({x}^{4} / 2\right) + C$

Jun 27, 2018

$\int {x}^{3} / \sqrt{4 + {x}^{8}} \mathrm{dx} = \frac{1}{4} \ln | {x}^{4} + \sqrt{{x}^{8} + 4} | + c$

#### Explanation:

Here,

$I = \int {x}^{3} / \sqrt{4 + {x}^{8}} \mathrm{dx} = \int {x}^{3} / \sqrt{4 + {\left({x}^{4}\right)}^{2}} \mathrm{dx}$

Subst. ${x}^{4} = u \implies 4 {x}^{3} \mathrm{dx} = \mathrm{du} \implies {x}^{3} \mathrm{dx} = \frac{1}{4} \mathrm{du}$

So,

$I = \int \frac{1}{\sqrt{4 + {u}^{2}}} \cdot \frac{1}{4} \mathrm{du}$

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

$= \frac{1}{4} \ln | u + \sqrt{{u}^{2} + {2}^{2}} | + c$

$= \frac{1}{4} \ln | {x}^{4} + \sqrt{{\left({x}^{4}\right)}^{2} + {2}^{2}} | + c$

$= \frac{1}{4} \ln | {x}^{4} + \sqrt{{x}^{8} + 4} | + c$