Hello The_Reporter,
We are given to compute the indefinite integral:
$$I=\int\frac{\sqrt{1-x^4}}{x^5}\,dx$$
Let's try the substitution:
$$x^2=\cos(\theta)\,\therefore\,dx=-\frac{\sin(\theta)}{2x}\,d\theta$$
And we obtain:
$$I=-\frac{1}{2}\int\frac{\sin^2(\theta)}{\cos^3(\theta)}\,d\theta$$
Next, let's try integration by parts where:
$$u=\sin(\theta)\,du=\cos(\theta)\,d\theta$$
$$dv=\frac{\sin(\theta)}{\cos(\theta}\,d\theta\,\therefore\,v=\frac{1}{2\cos^2(\theta)}$$
And now we obtain:
$$I=-\frac{1}{4}\left(\frac{\sin(\theta)}{\cos^2(\theta)}-\int\sec(\theta)\,d\theta\right)$$
For the remaining integral, consider:
$$\sec(\theta)=\frac{\sec(\theta)\left(\sec(\theta)+\tan(\theta)\right)}{\sec(\theta)+\tan(\theta)}=\frac{d}{d\theta}\left(\ln\left|\sec(\theta)+\tan(\theta)\right|\right)$$
Hence, we obtain:
$$I=-\frac{1}{4}\left(\frac{\sin(\theta)}{\cos^2(\theta)}-\ln\left|\sec(\theta)+\tan(\theta)\right|\right)+C$$
From our original substitution, we find that:
$$\tan(\theta)=\frac{\sqrt{1-x^4}}{x^2}$$
$$\sec(\theta)=\frac{1}{x^2}$$
And so, we finally find:
$$\bbox[5px,border:2px solid #207498]{I=\frac{1}{4}\left(\ln\left(\frac{\sqrt{1-x^4}+1}{x^2} \right)-\frac{\sqrt{1-x^4}}{x^4} \right)+C}$$