Solution of the equation ∫log⁡2x dxex−1=π6 are

Putting ex−1=t2, we have exdx=2t dtso ∫log⁡2x dxex−1=∫1ex−1 2tdttt2+1=2∫1ex−1 dtt2+1=2tan−1⁡ex−1−π4Thus the given equation reduces totan−1⁡ex−1−π4=π12⇒ex−1=tan⁡π3=3so ex=4⇒x=2log⁡2