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**Macleef** Integrate:

a) $\displaystyle \int \frac{2xln(x^2 + 1)}{x^2 + 1}\, dx

$

$\displaystyle Let\, u = x^2 + 1, du = 2x\, dx$

$\displaystyle = \int \frac{2xln(x^2 + 1)}{u} \frac{1}{2x}\, du$

$\displaystyle = \int ln(x^2 + 1) u^{-1}\, du$

$\displaystyle = ln(x^2 + 1) (ln(x^2 + 1)) + c$

$\displaystyle = [ln(x^2 + 1)]^2 + c$

answer is $\displaystyle \frac{1}{2} [ln(x^2 + 1)]^2 + c$, where does 1/2 come from??

b) $\displaystyle \int \frac {x}{2x + 1}\, dx = x(2x + 1)^{-1}\, dx$

$\displaystyle Let \, u = 2x + 1\, du = 2\, dx$

$\displaystyle = \frac{1}{2} \int x(u^{-1})\, du$

$\displaystyle = (\frac{1}{2})(\frac{1}{2})(x^2)(ln(2x + 1)) + c$

$\displaystyle = \frac{1}{4}x^2 (ln|2x + 1|) + c$

answer is $\displaystyle \frac{1}{2}x - \frac{1}{4}(ln|2x + 1|) + c$, what did I do wrong?