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Question

log50exex1ex+3dx=
  1. 4π
  2. 2+π
  3. π4
  4. 3+2π

A
3+2π
B
π4
C
4π
D
2+π
Solution
Verified by Toppr

Let I=exex1ex+3
Substitute t=exdt=exdx, we get
I=t1t+3dt
Substitute u=t1du=12t1dt, we get
I=2u2u2+4du=2(14u2+4)du=2du24u2+4du=2u4tan1(u2)=2t14tan1(t12)=2ex14tan1(ex12)
Therefore,
log50Idx=[2ex14tan1(ex12)]log50=4π

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