What is f(x) if f'(x) = 2x/x^2-1 and f(2) = 0?

Answer 1

If

f'(x) = 2x/x^2-1

mean

f'(x) = 2 x / ( x² -1 )

then

f (x) = ∫ f'(x) dx = ∫ 2 x dx / ( x² -1 )
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Remark:

Substitution

x² -1 = u

2 x dx = du
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f (x) = ∫ du / u = ℓn ( u ) + C

f (x) = ℓn ( x² -1 ) + C

f(2) = 0

f (2) = ℓn ( 2² -1 ) + C = 0

ℓn ( 4 -1 ) + C = 0

ℓn ( 3 ) + C = 0

Subtract ℓn ( 3 ) to both sides

ℓn ( 3 ) + C - ℓn ( 3 ) = 0 - ℓn ( 3 )

C = - ℓn ( 3 )

f (x) = ℓn ( x² -1 ) + C

f (x) = ℓn ( x² -1 ) - ℓn ( 3 )

f (x) = ℓn [ ( x² -1 ) / 3 ]
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Remark:

ℓn ( a ) - ℓn ( b) = ℓn ( a / b )

so

ℓn ( x² -1 ) - ℓn ( 3 ) = ℓn [ ( x² -1 ) / 3 ]
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ℓn mean natural logarithm ( logarithm to the base e )

Answer 2

Thank you, Bosnian brother.
That is exactly what I put on my test, I was just making sure.