Answer
See below.
Work Step by Step
Differentiate the RHS and see if it equals$\quad \ln(a^{2}+x^{2})$
$\displaystyle \frac{d}{dx}[x\ln(a^{2}+x^{2})]=(1)\ln(a^{2}+x^{2})+x\cdot(\frac{1}{a^{2}+x^{2}}\cdot 2x)$
$=\displaystyle \ln(a^{2}+x^{2})+\frac{2x^{2}}{a^{2}+x^{2}}$
$\displaystyle \frac{d}{dx}[2x]=2$
$\displaystyle \frac{d}{dx}[2a\tan^{-1}(\frac{x}{a})]=2a\cdot\frac{1}{1+(\frac{x}{a})^{2}}\cdot\frac{1}{a}=\frac{2}{\frac{a^{2}+x^{2}}{a^{2}}}=\frac{2a^{2}}{a^{2}+x^{2}}$
$\displaystyle \frac{d}{dx}[RHS]=\ln(a^{2}+x^{2})+\frac{2x^{2}}{a^{2}+x^{2}}-2+\frac{2a^{2}}{a^{2}+x^{2}}$
$=\displaystyle \ln(a^{2}+x^{2})+\frac{2x^{2}}{a^{2}+x^{2}}+\frac{-2(a^{2}+x^{2})+2a^{2}}{a^{2}+x^{2}}$
$=\displaystyle \ln(a^{2}+x^{2})+\frac{2x^{2}}{a^{2}+x^{2}}-\frac{2x^{2}}{a^{2}+x^{2}}$
$=\ln(a^{2}+x^{2})$
which verifies the formula, as needed.
