If ∫dx(x2−1)√x=p⋅ln∣∣∣f(x)−1f(x)+1∣∣∣+q⋅tan−1(g(x))+C.
Then the value of f(4)p−q⋅g(1)=
(Where p,q are fixed constants and C is integration constant)
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Solution
Let I=∫dx(x2−1)√x
Put x=t2,dx=2tdt
The integral becomes : I=∫2dtt4−1=∫[1t2−1−1t2+1]dt=12ln∣∣∣t−1t+1∣∣∣−tan−1t+C=12ln∣∣∣√x−1√x+1∣∣∣−tan−1√x+C ⇒p=12,q=−1,f(x)=√x,g(x)=√x
So, f(4)p−q⋅g(1)=5