∫log⁡2k dxex−1=π6 then k=

∫log⁡2k exdxex⋅ex−1  put ex−1=t∫1e−1 2tdtt2+1t          ex−1=t2exdx=2tdt2∫1eε−1 dt1+t2=2tan−1⁡t1et−1                  LLt=2−1t=1∣ULt=ek−1⇒2tan−1⁡ek−1−π4=π62tan−1⁡ek−1=π6+π2tan−1⁡ek−1=π3ek−1=3ek=4k=log⁡4