# Integral 01

• Aug 17th 2008, 04:54 PM
Apprentice123
Integral 01
$\int \frac{dx}{x^4\sqrt{1+x^2}}$

$\frac{(2x^2-1)(1+x^2)\frac{1}{2}}{3x^3}$
• Aug 17th 2008, 05:24 PM
Chop Suey
$\int \frac{dx}{x^4\sqrt{1+x^2}}$

You can use trigonometric subtitution.

Let:
$x = \tan{\theta}$

$dx = \sec^2{\theta} d\theta$

$\int \frac{\sec^2{\theta}}{(\tan{\theta})^4\sqrt{1+\tan ^2{\theta}}} d\theta$

$\int \frac{\sec^2{\theta}}{(\tan{\theta})^4\sec{\theta} } d\theta$

$\int \frac{\sec{\theta}}{(\tan{\theta})^4} d\theta$

$\int \frac{1}{\cos{\theta}} \cdot \frac{cos^4{\theta}}{\sin^4{\theta}} d\theta$

$\int \frac{cos^3{\theta}}{\sin^4{\theta}} d\theta$

$\int \frac{(1-sin^2{\theta})\cos{\theta}}{\sin^4{\theta}} d\theta$

Let
$u = \sin{\theta}$

$du = \cos{\theta} d\theta$

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

Can you continue?
• Aug 17th 2008, 07:21 PM
Krizalid
Think in a reciprocal substitution, like $u=\frac1x,$ the integral equals $-\int{\frac{u^{3}}{\sqrt{1+u^{2}}}\,du}.$ Now put $z^2=1+u^2.$
• Aug 18th 2008, 08:04 AM
Apprentice123
Look my solution

$\int \frac{dx}{x^4\sqrt{1+x^2}} = \int x^(-4) (1+x^2)^-\frac{1}{2}dx$

$1+x^2=z^2$
$x^2=z^2-1$
$dx= \frac{zdz}{\sqrt{z^2-1}}$

$\int (z^2+1)^(-2) (z^2)^\frac{1}{2}\frac{z}{\sqrt{z^2-1}}$