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Question

If dx(x21)x=plnf(x)1f(x)+1+qtan1(g(x))+C.
Then the value of f(4)pqg(1)=
(Where p,q are fixed constants and C is integration constant)

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Solution

Let I=dx(x21)x
Put x=t2,dx=2tdt
The integral becomes :
I=2dtt41=[1t211t2+1]dt =12lnt1t+1tan1t+C =12lnx1x+1tan1x+C
p=12,q=1,f(x)=x,g(x)=x
So, f(4)pqg(1)=5

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