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Question

If dx(x+100)x+99=f(x)+C, then f(x) is equal to (where C is integration constant)

A
tan1(x+99)
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B
tan1(3x+99)
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C
2tan1(x+99)
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D
2sin1(x+99)
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Solution

The correct option is C 2tan1(x+99)
Let, I=dx(x+100)x+99
Put x+99=t2dx=2t dt
I=2t dt(t2+1)t
=21t2+1dt
=2tan1t+C
=2tan1(x+99)+C
f(x)=2tan1(x+99)

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