1. integration problem no.3

i stop here because the substitution method doesn't work

2. Hello, afeasfaerw23231233!

With a quadratic form: . $\displaystyle x^2+a^2,\;\;x^2-a^2,\;\;a^2-x^2$
. . a Trigonometric Substitution is recommended.

Find $\displaystyle \int\frac{\sqrt{1-x^2}}{x^4}\,dx$

Let $\displaystyle x \:=\:\sin\theta\quad\Rightarrow\quad dx\:=\:\cos\theta\,d\theta$
. . and: .$\displaystyle \sqrt{1-x^2} \:=\:\sqrt{1-\sin^2\!\theta} \:=\:\sqrt{\cos^2\!\theta} \:=\:\cos\theta$

Substitute: .$\displaystyle \int\frac{\cos\theta}{\sin^4\!\theta}(\cos\theta\, d\theta) \;=\;\int\frac{\cos^2\!\theta}{\sin^2\!\theta}\cdo t\frac{1}{\sin^2\!\theta}\,d\theta \;=\;\int\cot^2\!\theta\csc^2\!\theta\,d\theta$

Let $\displaystyle u\,=\,\cot\theta\quad\Rightarrow\quad du \,=\,-\csc^2\!\theta\,d\theta$

Substitute: .$\displaystyle -\int u^2\,du \;=\;-\frac{1}{3}u^3 + C$

Back-substitute: .$\displaystyle -\frac{1}{3}\cot^3\!\theta + C$ .[1]

Back-substitute again. .We have: .$\displaystyle \sin\theta \:= \:\frac{x}{1} \:= \:\frac{opp}{hyp}$

$\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = x,\;hyp = 1$
Using Pythagorus, we have: .$\displaystyle adj = \sqrt{1-x^2}$
. . Hence: .$\displaystyle \cot\theta \:=\:\frac{adj}{opp} \:=\:\frac{\sqrt{1-x^2}}{x}$

Substitute into [1]: .$\displaystyle -\frac{1}{3}\left(\frac{\sqrt{1-x^2}}{x}\right)^3 + C \;=\;\boxed{-\frac{(1-x^2)^{\frac{3}{2}}}{3x^3}}$

3. $\displaystyle \int {\frac{{\sqrt {1 - x^2 } }} {{x^4 }}\,dx}$

Define $\displaystyle x=\frac1u,$ the integral becomes to

$\displaystyle - \int {u\sqrt {u^2 - 1} \,du} = - \frac{{\left( {u^2 - 1} \right)^{3/2} }} {3} + k$

Back substitute and we're done.