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This is my solution of this problem :

Question. $\displaystyle \int \frac{x^3}{1+x^2} \, \mathrm{d}x$

Solution. Let $1+x^2 = u$. Then

$$\frac{\mathrm{d}u}{\mathrm{d}x} = 2x \quad\Rightarrow \quad \mathrm{d}u = 2x \, \mathrm{d}x \quad \Rightarrow \quad x \, \mathrm{d}x = \frac{1}{2}\, \mathrm{d}u. $$

Also, $x^2 = u - 1$. Using both, we get

\begin{align*} \int \frac{x^3}{1+x^2} \, \mathrm{d}x &= \int \frac{x \cdot x^2}{1+x^2} \, \mathrm{d}x = \int \frac{u-1}{u} \, \mathrm{d}u \\ &= \int \left( 1 - \frac{1}{u} \right) \, \mathrm{d}u = u - \log u + \mathsf{C} \\ &= 1 + x^2 - \log (1 + x^2) + \mathsf{C} \end{align*}

But the answer is this

$$ \frac{x^2 - \log(x^2 + 1)}{2} + \mathsf{C}. $$

What am I doing wrong in my solution ?

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    $\begingroup$ You lost the 1/2 from the substitution $x \ dx = \frac{1}{2} \ du$. Then absorb the extra additive $1/2$ into the "$+C$". $\endgroup$
    – Randall
    Apr 11 '19 at 17:24
  • $\begingroup$ You are right.. $\endgroup$
    – arandomguy
    Apr 11 '19 at 17:27
  • $\begingroup$ @Randall Why are you posting an answer in the comment section? $\endgroup$
    – Arthur
    Apr 11 '19 at 17:27
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  1. You forgot the $\frac12$ in $x\,dx=\frac12\,du$
  2. The $1$ (which becomes $\frac12$ after taking the above point into account) can be absorbed into the $C$
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