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Question Number 125426 by joki last updated on 11/Dec/20

Answered by bramlexs22 last updated on 11/Dec/20

$$x^2-1=u$$$$\frac{1}{2}\int \frac{du}{u^2(u+2)}=\frac{1}{2}\int [\frac{A}{u}+\frac{B}{u^2}+\frac{C}{u+2}]du$$$$=\frac{1}{2}(-\frac{1}{4}\ln \mid u\mid -\frac{1}{2u}+\frac{1}{4}\ln \mid u+2\mid )+c$$$$=-\frac{1}{8}\ln \mid u\mid -\frac{1}{4u}+\frac{1}{8}\ln \mid u+2\mid +c$$$$=\frac{1}{8}\ln \mid \frac{x^2+1}{x^2-1}\mid -\frac{1}{4(x^2-1)}+c$$

Commented by joki last updated on 11/Dec/20

$$ldontundestandsir,pleasehelpmeexplain$$$$withdetail.thankssir$$

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