# How do you integrate int x(x^2+3)dx?

$\int x \left({x}^{2} + 3\right) \mathrm{dx} = \frac{1}{4} {\left({x}^{2} + 3\right)}^{2} + C$
It can be done by other methods, but if you recognise that the $x$ outside the bracket results from the bracket differentiated we can integrate by inspection.
now $\frac{d}{\mathrm{dx}} \left({\left({x}^{2} + 3\right)}^{2}\right) = 2 \times 2 x \left({x}^{2} + 3\right) = 4 x \left({x}^{2} + 3\right)$
Thus $\int x \left({x}^{2} + 3\right) \mathrm{dx} = \frac{1}{4} {\left({x}^{2} + 3\right)}^{2} + C$