I=∫αα+1(x+α)(x+α+1)dx=∫αα+1(x+α1–x+α+11)dx
Using partial fractions
=[ln∣x+α∣−ln∣x+α+1∣]αα+1
=[ln∣x+α+1x+α∣]αα+1
=ln∣2α+22α+1∣−ln∣2α+12α∣
=ln∣(2α+1)2−1(2α+1)2∣=ln89
⇒(2α+1)2=9
⇒2α+1=±3
⇒α=1 or −2.
A value of α such that ∫αα+1(x+α)(x+α+1)dx=loge(89) is
Held on 12 Apr 2019 · Verified 6 Jul 2026.
−21
21
−2
2
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