I am trying to calculate the integral of the form $$ \begin{align*} \frac{\Gamma(n+1/2)(2r)^n}{\sqrt{\pi} \lambda^n r} \int_0^\infty dQ \frac{\lambda^{2n}\sin(Qr)}{(\lambda^2 + r^2)^{n+1}}Q \end{align*} $$
An idea was to use Basset's integral to solve it, $$ \begin{align*} K_n(\lambda r)=\frac{\Gamma(n+1/2)(2r)^n}{\sqrt{\pi} \lambda^n} \int_0^\infty dQ \frac{\cos(Qr)}{(\lambda^2 + r^2)^{n+1}} \end{align*} $$ Differentiating under the integral sign is the way I believe. How to differential the modified Bessel's function of second kind $K_n(\lambda r)$ with respect to $r$.
You have the right idea. Let $f(r)$ be given by the initegral
$$\begin{align} f(r)&=\int_0^\infty \frac{\sin(Qr)}{Q(\lambda^2+Q^2)^{n+1}}\,dQ\\\\ \end{align}$$
Owing to the uniform convergence of the integral of the differentiated integrand, we can differentiate under the integral. Proceeding and using Basset's integral we find that
$$\begin{align} f'(r)&=\int_0^\infty \frac{\cos(Qr)}{(\lambda^2+Q^2)^{n+1}}\,dQ\\\\ &=\frac{\pi^{1/2} r^{n+1/2}}{n!(2\lambda)^{n+1/2}}K_{n+1/2}(\lambda r) \end{align}$$