square root integral

[intervade]

Hello,

I'm having a problem with this integral. Its almost like it could be a basic u, du substitution but I'm not really sure where to start.

$\displaystyle \int \frac{1}{x^3\sqrt{1+x^4}}dx$

Some direction on this would be much appreciated!

 

[BigGlenntheHeavy]

$\displaystyle \int\frac{dx}{x^{3}(1+x^{4})^{1/2}}, \ Let \ u \ = \ (1+x^{4})^{1/2}, \ du \ = \ \frac{1}{2}(1+x^{4})^{-1/2}(4x^{3})dx$

$\displaystyle du \ = \ \frac{2x^{3}}{u}dx, \ dx \ = \ \frac{udu}{2x^{3}}$

$\displaystyle Also \ note, \ 1+x^{4} \ = \ u^{2}, \ x^{4} \ = \ u^{2}-1, \ x^{6} \ = \ (u^{2}-1)^{3/2}$

$\displaystyle Hence, \ \frac{1}{2}\int\frac{udu}{x^{6}u} \ = \ \frac{1}{2}\int\frac{du}{x^{6}} \ = \ \frac{1}{2}\int\frac{du}{(u^{2}-1)^{3/2}} \ = \ \ - \frac{u}{2(u^{2}-1)^{1/2}} \ +C$

$\displaystyle Re- Subbing, \ we \ get \ = \ - \ \frac{(1+x^{4})^{1/2}}{2(x^{4})^{1/2}}+C \ = \ -\frac{(1+x^{4})^{1/2}}{2x^{2}}+C.$

$\displaystyle Check: \ D_x\bigg[-\frac{(1+x^{4})^{1/2}}{2x^{2}}+C\bigg] \ = \ \frac{1}{x^{3}(1+x^{4})^{1/2}}.$

 

[BigGlenntheHeavy]

$\displaystyle This \ is \ probably \ the \ easiest \ way, \ to \ wit:$

$\displaystyle \int\frac{dx}{x^{3}(1+x^{4})^{1/2}} \ = \ \int\frac{dx}{x^{3}[1+(x^{2})^{2}]^{1/2}}$

$\displaystyle Now, \ let \ x^{2} \ = \ tan(\theta), \ \implies \ x \ = \ tan^{1/2}(\theta) \ and \ x^{3} \ = \ tan^{3/2}(\theta)$

$\displaystyle Also, \ 2xdx \ = \ sec^{2}(\theta)d\theta, \ dx \ = \ \frac{sec^{2}(\theta)d\theta}{2x} \ = \ \frac{sec^{2}(\theta)d\theta}{2tan^{1/2}(\theta)}$

$\displaystyle Hence, \ we \ have: \ \int\frac{dx}{x^{3}[1+(x^{2})^{2}]^{1/2}} \ = \ \frac{1}{2}\int\frac{sec^{2}(\theta)d\theta}{tan^{1/2}(\theta)tan^{3/2}(\theta)[1+tan^{2}(\theta)]^{1/2}}$

$\displaystyle = \ \frac{1}{2}\int\frac{sec^{2}(\theta)d\theta}{tan^{2}(\theta)sec(\theta)} \ = \ \frac{1}{2}\int\frac{sec(\theta)d\theta}{tan^{2}(\theta)} \ = \ \frac{1}{2}\int\frac{cos(\theta)d\theta}{sin^{2}(\theta)},$

$\displaystyle = \ \frac{1}{2}\int cot(\theta)csc(\theta)d\theta \ = \ -\frac{1}{2}csc(\theta)+C \ = \ -\frac{1}{2}\bigg[\frac{(x^{4}+1)^{1/2}}{x^{2}}\bigg]+C \ = \ -\frac{(x^{4}+1)^{1/2}}{2x^{2}}+C$