All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 119335 by physicstutes last updated on 23/Oct/20
Expressf(x)=1(x−1)2(x2+1)intopartialfractions.henceevaluateI=∫40f(x)dx
Answered by floor(10²Eta[1]) last updated on 23/Oct/20
1(x−1)2(x2+1)=Ax−1+B(x−1)2+Cx+Dx2+11=A(x−1)(x2+1)+B(x2+1)+(Cx+D)(x−1)2=(A+C)x3−(A−B+2C+D)x2+(A+C−2D)x−A+B+D⇒A+C=0A−B+2C+D=0A+C−2D=0−A+B+D=1⇒D=0⇒C=1/2⇒A=−1/2⇒B=1/21(x−1)2(x2+1)=−12(x−1)+12(x−1)2+x2(x2+1)
Answered by mathmax by abdo last updated on 24/Oct/20
f(x)=1(x−1)2(x2+1)⇒f(x)=ax−1+b(x−1)2+cx+dx2+1b=12,limx→+∞xf(x)=0=a+c⇒c=−a⇒f(x)=ax−1+12(x−1)2+−ax+dx2+1f(0)=1=−a+12+d⇒−a+d=12⇒a−d=−12f(2)=15=a+12+−2a+d5⇒1=5a+52−2a+d⇒3a+d=1−52=−32wehavea=d−12⇒3(d−12)+d=−32⇒4d=0⇒d=0⇒a=−12⇒f(x)=−12(x−1)+12(x−1)2+x2(x2+1)so∫04f(x)dx=−12∫04dxx−1+12∫04dx(x−1)2+14∫042xx2+1dx=−12[ln∣x−1∣]04−12[1x−1]04+14[ln(x2+1)]04=−12ln(3)−12(13+1)+14ln(17)=−ln32−23+ln174
Terms of Service
Privacy Policy
Contact: info@tinkutara.com