All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 128221 by john_santu last updated on 05/Jan/21
Givenf(x+1x)=x4−1x4+2then∫12(1−x−2)f(x)dx=
Answered by liberty last updated on 05/Jan/21
x4−1x4+2=(x2+1x2)(x2−1x2)+2=[(x+1x)2−2](x+1x)(x−1x)+2(∙)x−1x=(x−1x)2=(x+1x)2−4thenf(x+1x)=[(x+1x)2−2](x+1x)(x+1x)2−4+2orf(x)=(x2−2)xx2−4+2then∫12(x2−1x2)[x(x2−2)x2−4+2]dx=∫12(x2−1)(x2−2)x2−4xdx−∫122(x2−1)x2dxI1=∫12(x4−3x2+2)x2−4xdxI2=∫122(1−x−2)dx
Terms of Service
Privacy Policy
Contact: info@tinkutara.com