All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 84702 by Power last updated on 15/Mar/20
Commented by abdomathmax last updated on 15/Mar/20
I=∫dxshx+1⇒I=∫dxex−e−x2+1=∫2dxex−e−x+2=ex=t∫2t−t−1+2dtt=∫2dtt2−1+2t=∫2dtt2+2t−1=∫2dt(t+1)2−2=∫2dt(t+1−2)(t+1+2)=222∫(1t+1−2−1t+1+2)dt=12ln∣t+1−2t+1+2∣+C=12ln∣ex+1−2ex+1+2∣+C
Commented by Power last updated on 15/Mar/20
thanks
youarewelcome
Terms of Service
Privacy Policy
Contact: info@tinkutara.com