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Question Number 196375 by RoseAli last updated on 23/Aug/23

$$\int \frac{dx}{x(x^4-1)}$$

Commented by RoseAli last updated on 23/Aug/23

$$\int \frac{dx}{x(x^4+1)}$$

Answered by MM42 last updated on 23/Aug/23

$$a)\frac{1}{4}ln\left(\frac{x^4-1}{x^4}\right)+c✓$$$$b)\frac{1}{4}ln\left(\frac{x^4}{x^4+1}\right)+c✓$$$$$$

Commented by RoseAli last updated on 23/Aug/23

$$\mathrm{step}\mathrm{by}\mathrm{step}$$

Commented by Frix last updated on 23/Aug/23

$$I=\int \frac{dt}{x(x^4+a)}$$$$\mathrm{Simply}\mathrm{use}t=x^4\Rightarrow$$$$I=\frac{\ln \mid x\mid }{a}-\frac{\ln \mid x^4+a\mid }{4a}+C$$

Commented by MM42 last updated on 24/Aug/23

$$a)\frac{1}{x(x^4-1)}=\frac{a}{x}+\frac{b}{x-1}+\frac{c}{x+1}+\frac{dx+e}{x^2+1}$$$$\Rightarrow a=-1,b=\frac{1}{4},c=\frac{1}{4},d=\frac{1}{2},e=0$$$$\Rightarrow I=\int (-\frac{1}{x}+\frac{1}{4(x-1)}+\frac{1}{4(x+1)}+\frac{x}{2(x^2+1)})dx$$$$=-lnx+\frac{1}{4}ln(x-1)+\frac{1}{4}ln(x+1)+\frac{1}{4}ln(x^2+1)+c$$$$=\frac{1}{4}ln\left(\frac{x^4-1}{x^4}\right)+c$$$$or$$$$\frac{1}{x(x^4-1)}=-\frac{x^4-1+x^4}{x(x^4-1)}=-\frac{1}{x}+\frac{x^3}{x^4-1}\to ....$$$$b)\frac{1}{x(x^4+1)}=\frac{x^4+1-x^4}{x(x^4+1)}=\frac{1}{x}-\frac{x^3}{x^4+1}$$$$\Rightarrow I=lnx-\frac{1}{4}ln(x^4+1)+c$$$$=\frac{1}{4}ln\frac{x^4+1}{x^4}+c$$

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