Question Number 125426 by joki last updated on 11/Dec/20
Answered by bramlexs22 last updated on 11/Dec/20
![x^2 −1= u (1/2)∫ (du/(u^2 (u+2))) = (1/2)∫ [(A/u)+(B/u^2 )+(C/(u+2))] du = (1/2)(−(1/4)ln ∣u∣ −(1/(2u))+(1/4)ln ∣u+2∣)+c =−(1/8)ln ∣u∣−(1/(4u)) + (1/8)ln ∣u+2∣ + c =(1/8)ln ∣((x^2 +1)/(x^2 −1)) ∣−(1/(4(x^2 −1))) + c](https://www.tinkutara.com/question/Q125430.png)
$${x}^{\mathrm{2}} \:−\mathrm{1}=\:{u} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{du}}{{u}^{\mathrm{2}} \left({u}+\mathrm{2}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\left[\frac{{A}}{{u}}+\frac{{B}}{{u}^{\mathrm{2}} }+\frac{{C}}{{u}+\mathrm{2}}\right]\:{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid{u}\mid\:−\frac{\mathrm{1}}{\mathrm{2}{u}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid{u}+\mathrm{2}\mid\right)+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid{u}\mid−\frac{\mathrm{1}}{\mathrm{4}{u}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid{u}+\mathrm{2}\mid\:+\:{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\mid−\frac{\mathrm{1}}{\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:+\:{c} \\ $$
Commented by joki last updated on 11/Dec/20
$${l}\:{dont}\:{undestand}\:{sir},{please}\:{help}\:{me}\:{explain}\: \\ $$$${with}\:{detail}.{thanks}\:{sir} \\ $$