$\displaystyle \int \frac{dx}{x^4\sqrt{1+x^2}}$

answer:

$\displaystyle \frac{(2x^2-1)(1+x^2)^\frac{1}{2}}{3x^3}$

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- Aug 6th 2008, 05:06 AMApprentice123Integral4
$\displaystyle \int \frac{dx}{x^4\sqrt{1+x^2}}$

answer:

$\displaystyle \frac{(2x^2-1)(1+x^2)^\frac{1}{2}}{3x^3}$ - Aug 6th 2008, 06:20 AMgalactus
There are many ways, but I like trig sub for some reason.

Let $\displaystyle x=tan(u), \;\ dx=sec^{2}(u)du$

When we make the subs we get:

$\displaystyle \int csc(u)cot^{3}(u)du$

$\displaystyle =\int csc(u)(csc^{2}(u)-1)cot(u)du$

Now, can you continue?. Another sub should do it, and when you get to the end remember to let $\displaystyle u=tan^{-1}(x)$ to get it back in terms of x. - Aug 6th 2008, 06:27 AMkalagota
also, from same substitution,

$\displaystyle \int \frac{\cos^3 u}{\sin^4 u} \, du$

$\displaystyle t= \sin u$

$\displaystyle dt = \cos u \, du$

$\displaystyle \cos^2 u = 1-\sin^2 u = 1-t^2$

$\displaystyle \int \frac{1-t^2}{t^4} \, dt$ - Aug 6th 2008, 11:05 AMKrizalid
A reciprocal substitution also works.

- Aug 6th 2008, 04:06 PMApprentice123
thanks

- Aug 6th 2008, 06:39 PMmr fantastic
- Aug 7th 2008, 11:14 AMKrizalid
Dunno. I said it 'cause I saw nobody told it before.