I am unsure on how to solve this integral
$$\displaystyle\int x\cdot\ln(x^2+3) ·x dx.$$
I thought of multiplying $x$ by $x$ in order to get $x²$ so that the equation would become $x²\ln(x^2+3)dx.$
However, I am not sure how to proceed from here.
Any help?
On
In terms of integrating the rational function part only in @GEdgar's solution we have I think (missing out constant of integration)
$$\int (a x^2-b) \; dx=\frac{ax^3}{3}-bx=\int \frac{(ax^2-b)(ax^2+b)}{ax^2+b} \; dx$$
which can be rearranged to give $$a^2 \int \frac{x^4}{ax^2+b} \; dx=\frac{ax^3}{3}-bx+b^2\int \frac{1}{ax^2+b} \; dx$$
Since $a=1$ in this example we have $$\int \frac{x^4}{x^2+b} \; dx=\frac{x^3}{3}-bx+b^2\int \frac{1}{x^2+b} \; dx$$
and since $\frac{d}{dx}$($\arctan(x/\sqrt{b}))=\frac{\sqrt{b}}{x^2+b}$
$$\int \frac{x^4}{x^2+b} \; dx=\frac{x^3}{3}-bx+\frac{b^2}{\sqrt{b}}\;\arctan(\frac{x}{\sqrt{b}})$$
Yes, the $x$'s may be multiplied so that we have $x^2$.
One way ... integrate by parts $$ \int \!{x}^{2}\ln \left( {x}^{2}+3 \right) \,{\rm d}x=\frac{{x}^{3} \ln \left( {x}^{2}+3 \right)}{3} -\frac{2}{3}\,\int \!{\frac {{x}^{4}}{{x}^{2}+3} }\,{\rm d}x $$ This is how you get rid of the log.
Presumably you know how to integrate a rational function. With that denominator: your answer involves an arctangent (or else complex numbers).
added
$$ u = \log(x^2+3),\qquad du= \frac{2x \;dx}{x^2+3} \\ dv=x^2\;dx,\qquad v = \frac{x^3}{3}, \\ uv = \frac{x^3\log(x^2+3)}{3} \\ \int u\;dv = \int x^2 \log(x^2+3)dx \\ \int v\;du = \int \frac{x^3}{3}\;\frac{2x \;dx}{x^2+3} = \frac{2}{3}\int \frac{x^4 \;dx}{x^2+3} $$