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Math Integration

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f (x)/(2x + 1) dx

let f represent integrate sign

let u = x, du = dx => dx = du

= f (u)/(2u + 1) du
= f u (2u + 1)^(-1) du
= (1/2)u^2 (ln|2u + 1|) + c
= (1/2)x^2 (ln|2x + 1|) + c

...what did I do wrong? The correct answer is

(1/2)x - (1/4)ln|2x + 1| + c

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