Find the integral ?

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Answer 1 · 2018-06-27T14:19:30

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

$\int \frac{x^{3} \cdot d x}{\sqrt{x^{8} + 4}}$ $int (x^3*dx)/(sqrt(x^8+4)$

= $\frac{1}{4} \int \frac{4 x^{3} \cdot d x}{\sqrt{{(x^{4})}^{2} + 4}}$ $1/4int (4x^3*dx)/(sqrt((x^4)^2+4)$

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

$\frac{1}{4} \int \frac{2 cosh u \cdot d u}{2 cosh u}$ $1/4int (2coshu*du)/(2coshu)$

= $\frac{1}{4} \int d u$ $1/4int du$

=1/4*u+C#

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

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

Answer 2 · 2018-06-27T14:28:16

$\int \frac{x^{3}}{\sqrt{4 + x^{8}}} d x = \frac{1}{4} ln ∣ ∣ x^{4} + \sqrt{x^{8} + 4} ∣ ∣ + c$ $intx^3/sqrt(4+x^8)dx=1/4ln|x^4+sqrt(x^8+4)|+c$

Here,

$I = \int \frac{x^{3}}{\sqrt{4 + x^{8}}} d x = \int \frac{x^{3}}{\sqrt{4 + {(x^{4})}^{2}}} d x$ $I=intx^3/sqrt(4+x^8)dx=intx^3/sqrt(4+(x^4)^2)dx$

Subst. $x^{4} = u \Rightarrow 4 x^{3} d x = d u \Rightarrow x^{3} d x = \frac{1}{4} d u$ $x^4=u=>4x^3dx=du=>x^3dx=1/4du$

So,

$I = \int \frac{1}{\sqrt{4 + u^{2}}} \cdot \frac{1}{4} d u$ $I=int1/sqrt(4+u^2)*1/4du$

$= \frac{1}{4} \int \frac{1}{\sqrt{u^{2} + 2^{2}}} d u$ $=1/4int1/sqrt(u^2+2^2)du$

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

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

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