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Question Number 211579 by MrGaster last updated on 13/Sep/24
∫(1/((1−x^4 )(√(1+x^2 ))))dx.
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{1}}{\left(\mathrm{1}−\boldsymbol{{x}}^{\mathrm{4}} \right)\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}\boldsymbol{{dx}}. \\ $$$$ \\ $$
Answered by lepuissantcedricjunior last updated on 13/Sep/24
∫(1/((1−x^2 )(1+x^2 )(√(1+x^2 ))))dx=k x=tant=>dx=(1+tan^2 t)dt k=∫((cost dt)/((1−tan^2 t)))=∫((cos^2 t)/(cos^2 t−sin^2 t))dt =∫((cos^2 t)/(cos2t))dt=∫((1+cos2t)/(2(cos2t)))dt =(1/2)t+∫(1/(2cos2t))dt o=tant=>dt=(do/(1+o^2 )) m=∫(dt/(2(2cos^2 t−1)))=∫(do/(2((2/(1+o^2 ))−1)×(1+o^2 ))) =∫(do/((4−2−2o^2 )))=∫(do/(2(1−o^2 ))) =(1/2)∫(do/((1−o)(1+o)))=(1/4)∫((1/(1−o))+(1/(1+o)))do =(1/4)ln∣((1+o)/(1−o))∣+c k=(1/2)arctant+(1/4)ln∣((1+arctant)/(1−arctant))∣+c k=(1/2)arctan(arctant)+(1/4)ln∣((1+arctan(arctanx))/(1−arctan(arctanx)))∣+c
$$\int\frac{\mathrm{1}}{\left(\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} \right)\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}\boldsymbol{{dx}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{x}}=\boldsymbol{{tant}}=>\boldsymbol{{dx}}=\left(\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{t}}\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{k}}=\int\frac{\boldsymbol{{cost}}\:\:\boldsymbol{{dt}}}{\left(\mathrm{1}−\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{t}}\right)}=\int\frac{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{t}}}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{t}}−\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{t}}}\boldsymbol{{dt}} \\ $$$$\:\:\:=\int\frac{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{t}}}{\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{t}}}\boldsymbol{{dt}}=\int\frac{\mathrm{1}+\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{t}}}{\mathrm{2}\left(\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{t}}\right)}\boldsymbol{{dt}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{t}}+\int\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{t}}}\boldsymbol{{dt}} \\ $$$$\boldsymbol{{o}}=\boldsymbol{{tant}}=>\boldsymbol{{dt}}=\frac{\boldsymbol{{do}}}{\mathrm{1}+\boldsymbol{{o}}^{\mathrm{2}} } \\ $$$$\:\:\:\boldsymbol{{m}}=\int\frac{\boldsymbol{{dt}}}{\mathrm{2}\left(\mathrm{2}\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{t}}−\mathrm{1}\right)}=\int\frac{\boldsymbol{{do}}}{\mathrm{2}\left(\frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{{o}}^{\mathrm{2}} }−\mathrm{1}\right)×\left(\mathrm{1}+\boldsymbol{{o}}^{\mathrm{2}} \right)} \\ $$$$\:\:\:=\int\frac{\boldsymbol{{do}}}{\left(\mathrm{4}−\mathrm{2}−\mathrm{2}\boldsymbol{{o}}^{\mathrm{2}} \right)}=\int\frac{\boldsymbol{{do}}}{\mathrm{2}\left(\mathrm{1}−\boldsymbol{{o}}^{\mathrm{2}} \right)} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\boldsymbol{{do}}}{\left(\mathrm{1}−\boldsymbol{{o}}\right)\left(\mathrm{1}+\boldsymbol{{o}}\right)}=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{{o}}}+\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{o}}}\right)\boldsymbol{{do}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{ln}}\mid\frac{\mathrm{1}+\boldsymbol{{o}}}{\mathrm{1}−\boldsymbol{{o}}}\mid+\boldsymbol{{c}} \\ $$$$\:\:\boldsymbol{{k}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{arctant}}+\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{ln}}\mid\frac{\mathrm{1}+\boldsymbol{{arctant}}}{\mathrm{1}−\boldsymbol{{arctant}}}\mid+{c} \\ $$$$\boldsymbol{{k}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{arctan}}\left(\boldsymbol{{arctant}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{ln}}\mid\frac{\mathrm{1}+\boldsymbol{{arctan}}\left(\boldsymbol{{arctanx}}\right)}{\mathrm{1}−\boldsymbol{{arctan}}\left(\boldsymbol{{arctanx}}\right)}\mid+\boldsymbol{{c}} \\ $$$$ \\ $$
Commented by Frix last updated on 13/Sep/24
3^(rd) line: x=tan t ∧ dx=(1+tan^2 t)dt leads to k=∫((cos^3 t)/(cos^2 t −sin^2 t))dt ⇒ everything else is wrong.
$$\mathrm{3}^{\mathrm{rd}} \:\mathrm{line}: \\ $$$${x}=\mathrm{tan}\:{t}\:\wedge\:{dx}=\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{t}\right){dt} \\ $$$$\mathrm{leads}\:\mathrm{to} \\ $$$${k}=\int\frac{\mathrm{cos}^{\mathrm{3}} \:{t}}{\mathrm{cos}^{\mathrm{2}} \:{t}\:−\mathrm{sin}^{\mathrm{2}} \:{t}}{dt} \\ $$$$\Rightarrow\:\mathrm{everything}\:\mathrm{else}\:\mathrm{is}\:\mathrm{wrong}. \\ $$
Answered by Frix last updated on 21/Sep/24
∫(dx/((1−x^4 )(√(1+x^2 )))) =^([t=(((√2)x)/( (√(x^2 +1))))]) =((√2)/4)∫ ((t^2 −2)/(t^2 −1))dt=((√2)/8)∫((1/(t+1))−(1/(t−1))+2)dt= =((√2)/( 8))(2t+ln ((t+1)/(t−1)))= =(x/(2(√(x^2 +1))))+((√2)/8)ln ∣(((√2)x+(√(x^2 +1)))/( (√2)x−(√(x^3 +1))))∣ +C
$$\int\frac{{dx}}{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\overset{\left[{t}=\frac{\sqrt{\mathrm{2}}{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right]} {=} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\:\frac{{t}^{\mathrm{2}} −\mathrm{2}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\int\left(\frac{\mathrm{1}}{{t}+\mathrm{1}}−\frac{\mathrm{1}}{{t}−\mathrm{1}}+\mathrm{2}\right){dt}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\:\mathrm{8}}\left(\mathrm{2}{t}+\mathrm{ln}\:\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\right)= \\ $$$$=\frac{{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{2}}{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{\mathrm{2}}{x}−\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}}\mid\:+{C} \\ $$

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