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Question Number 46221 by Meritguide1234 last updated on 22/Oct/18

Answered by MJS last updated on 22/Oct/18

I solved it but I′m too tired to type it

IsolveditbutI′mtootiredtotypeit

Commented by Meritguide1234 last updated on 23/Oct/18

post your two or three line approach

postyourtwoorthreelineapproach

Commented by MJS last updated on 23/Oct/18

will type completely when I′m wake enough

willtypecompletelywhenI′mwakeenough

Answered by MJS last updated on 23/Oct/18

∫((√(x^2 +1))/(x(√(x^4 +1))(√(x^8 −1))))dx=∫(dx/(x(x^4 +1)(√(x^2 −1))))= [t=(√(x^2 −1)) → dx=((√(x^2 −1))/x)dt] =∫(dt/((t^2 +1)(t^4 +2t^2 +2)))=∫(dt/(t^2 +1))−∫((t^2 +1)/(t^4 +2t^2 +2))dt ∫(dt/(t^2 +1))=arctan t =arctan (√(x^2 −1)) ∫((t^2 +1)/(t^4 +2t^2 +2))dt=∫((t^2 +1)/((t^2 −at+b)(t^2 +at+b)))dt= [a=(√(−2+2(√2))); b=(√2)] =(1/(2ab))∫(((b−1)t+a)/(t^2 −at+b))dt−(1/(2ab))∫(((b−1)t−a)/(t^2 +at+b))dt= =((b−1)/(4ab))∫((2t−a)/(t^2 −at+b))dt+((b+1)/(4b))∫(dt/(t^2 −at+b))−((b−1)/(4ab))∫((2t+a)/(t^2 +at+b))dt+((b+1)/(4b))∫(dt/(t^2 +at+b))= =((b−1)/(4ab))ln (t^2 −at+b) +((b+1)/(2b(√(4b−a^2 ))))arctan ((2t−a)/(√(4b−a^2 ))) −((b−1)/(4ab))ln (t^2 +at+b) +((b+1)/(2b(√(4b−a^2 ))))arctan ((2t+a)/(√(4b−a^2 )))= =((b−1)/(4ab))ln ∣((x^2 +b−1−a(√(x^2 −1)))/(x^2 +b−1+a(√(x^2 −1))))∣ +((b+1)/(2b(√(4b−a^2 ))))(arctan ((a+2(√(x^2 −1)))/(√(4b−a^2 ))) −arctan ((a−2(√(x^2 −1)))/(√(4b−a^2 )))) ∫((√(x^2 +1))/(x(√(x^4 +1))(√(x^8 −1))))dx= =((b−1)/(4ab))ln ∣((x^2 +b−1+a(√(x^2 −1)))/(x^2 +b−1−a(√(x^2 −1))))∣ +((b+1)/(2b(√(4b−a^2 ))))(arctan ((a−2(√(x^2 −1)))/(√(4b−a^2 ))) −arctan ((a+2(√(x^2 −1)))/(√(4b−a^2 ))))+arctan (√(x^2 −1)) +C with a=(√(−2+2(√2))); b=(√2)

∫x2+1xx4+1x8−1dx=∫dxx(x4+1)x2−1=[t=x2−1→dx=x2−1xdt]=∫dt(t2+1)(t4+2t2+2)=∫dtt2+1−∫t2+1t4+2t2+2dt∫dtt2+1=arctant=arctanx2−1∫t2+1t4+2t2+2dt=∫t2+1(t2−at+b)(t2+at+b)dt=[a=−2+22;b=2]=12ab∫(b−1)t+at2−at+bdt−12ab∫(b−1)t−at2+at+bdt==b−14ab∫2t−at2−at+bdt+b+14b∫dtt2−at+b−b−14ab∫2t+at2+at+bdt+b+14b∫dtt2+at+b==b−14abln(t2−at+b)+b+12b4b−a2arctan2t−a4b−a2−b−14abln(t2+at+b)+b+12b4b−a2arctan2t+a4b−a2==b−14abln∣x2+b−1−ax2−1x2+b−1+ax2−1∣+b+12b4b−a2(arctana+2x2−14b−a2−arctana−2x2−14b−a2)∫x2+1xx4+1x8−1dx==b−14abln∣x2+b−1+ax2−1x2+b−1−ax2−1∣+b+12b4b−a2(arctana−2x2−14b−a2−arctana+2x2−14b−a2)+arctanx2−1+Cwitha=−2+22;b=2

Commented by MJS last updated on 23/Oct/18

...please check, I already corrected some typos

...pleasecheck,Ialreadycorrectedsometypos

Commented by maxmathsup by imad last updated on 23/Oct/18

thank you sir you have played a criket match with this integral...

thankyousiryouhaveplayedacriketmatchwiththisintegral...

Commented by MJS last updated on 23/Oct/18

you′re welcome

you′rewelcome

Commented by Meritguide1234 last updated on 25/Oct/18

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