Differentiate $\int_{0}^{x}\frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\frac{x}{a}$ under the integral sign? Differentiate $\displaystyle \int_{0}^{x}\frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \dfrac{x}{a}$ under the integral sign to find the value of $$\int_{0}^{x} \frac{dx}{(x^2 + a^2)^2}$$

I am getting something $\frac{1}{2a^3}tan^{-1}\frac{x}{a}$, but not sure of it .

$$\begin{array}{rcl}
\displaystyle \int_{0}^{x}\frac{\mathrm dt}{t^2 + a^2}
&=&
\displaystyle \frac{1}{a}\tan^{-1}\frac{x}{a}
\\
\displaystyle \frac{\partial}{\partial(a^2)} \int_{0}^{x}\frac{\mathrm dt}{t^2 + a^2}
&=&
\displaystyle \frac{\partial}{\partial(a^2)} \left(\frac{1}{a}\tan^{-1}\frac{x}{a}\right)
\\
\displaystyle \int_{0}^{x}-\frac{\mathrm dt}{(t^2 + a^2)^2}
&=&
\displaystyle \frac{\partial a}{\partial(a^2)} \frac{\partial}{\partial a} \left(\frac{1}{a}\tan^{-1}\frac{x}{a}\right)
\\
\displaystyle \int_{0}^{x}\frac{\mathrm dt}{(t^2 + a^2)^2}
&=&
\displaystyle -\frac{1}{2a} \left(\frac{-1}{a^2}\tan^{-1}\frac{x}{a} + \frac1a \frac{1}{1+\left(\frac xa\right)^2} \frac{-x}{a^2}\right)
\\
&=&\displaystyle \frac{1}{2a} \left(\frac{1}{a^2}\tan^{-1}\frac{x}{a} + \frac{x}{a(x^2+a^2)}\right)\\
&=&\displaystyle \frac{1}{2a^3} \left(\tan^{-1}\frac{x}{a}+\frac {ax}{x^2+a^2}\right)\\
\end{array}$$

Due to a suggestion on using $\dfrac{\partial}{\partial a}$ instead of $\dfrac{\partial}{\partial(a^2)}$,
$$\begin{array}{rcl}
\displaystyle \int_{0}^{x}\frac{\mathrm dt}{t^2 + a^2}
&=&
\displaystyle \frac{1}{a}\tan^{-1}\frac{x}{a}
\\
\displaystyle \frac{\partial}{\partial a} \int_{0}^{x}\frac{\mathrm dt}{t^2 + a^2}
&=&
\displaystyle \frac{\partial}{\partial a} \left(\frac{1}{a}\tan^{-1}\frac{x}{a}\right)
\\
\displaystyle \int_{0}^{x}\frac{-2a\ \mathrm dt}{(t^2 + a^2)^2}
&=&
\displaystyle \frac{-1}{a^2}\tan^{-1}\frac{x}{a} + \frac1a \frac{1}{1+\left(\frac xa\right)^2} \frac{-x}{a^2}
\\
\displaystyle \int_{0}^{x}\frac{\mathrm dt}{(t^2 + a^2)^2}
&=&
\displaystyle -\frac{1}{2a} \left(\frac{-1}{a^2}\tan^{-1}\frac{x}{a} + \frac1a \frac{1}{1+\left(\frac xa\right)^2} \frac{-x}{a^2}\right)
\\
&=&\displaystyle \frac{1}{2a} \left(\frac{1}{a^2}\tan^{-1}\frac{x}{a} + \frac{x}{a(x^2+a^2)}\right)\\
&=&\displaystyle \frac{1}{2a^3} \left(\tan^{-1}\frac{x}{a}+\frac {ax}{x^2+a^2}\right)\\
\end{array}$$
