Why is my integration solution wrong?

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*}

$$\frac{x^2 - \log(x^2 + 1)}{2} + \mathsf{C}.$$
• You lost the 1/2 from the substitution $x \ dx = \frac{1}{2} \ du$. Then absorb the extra additive $1/2$ into the "$+C$". Apr 11 '19 at 17:24
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 $$C$$