We use the following claim:
Let $a,b$ two real numbers and $\{f_n\}$ a sequence of continuous functions on $\left[a,b\right]$ which converges uniformly to $f$ on $[a,b]$. Then
$$\lim_{n\to\infty}\int_a^bf_n(t)dt=\int_a^bf(t)dt.$$
Indeed, we have
$$\left|\int_a^bf_n(t)dt-\int_a^bf(t)dt\right|\leq (b-a)\sup_{a\leq x\leq b}|f_n(x)-f(x)|,$$
which converges to $0$ thanks to the uniform convergence on $[a,b]$.
In our case, we have
$$\int_0^{\frac{\pi}2}f_n(x)dx=n\left[-\frac{(\cos x)^{n+1}}{n+1}\right]_0^{\frac \pi 2}=\frac n{n+1}\to 1,$$
whereas $f_n$ converges pointwise to $0$. This shows that the convergence cannot be uniform.
Since the integrand has a singularity at $x=0$, you must split the integral into two, such as
$$
\int_{0}^{\infty }\frac{1}{\sqrt{x^{3}+x}}dx=\int_{0}^{b}\frac{1}{\sqrt{x^{3}+x}}dx+\int_{b}^{\infty }\frac{1}{\sqrt{x^{3}+x}}dx\qquad b>0.
$$
As for the first integral use the limit test
$$\lim_{x\rightarrow 0}\frac{\frac{1}{\sqrt{x^{3}+x}}}{\frac{1}{\sqrt{x}}}
=\lim_{x\rightarrow 0}\frac{1}{\sqrt{x^{2}+1}}=1$$
to conclude that $\int_{0}^{b}\frac{1}{\sqrt{x^{3}+x}}dx$ is convergent,
because so is $\int_{0}^{b}\frac{1}{\sqrt{x}}dx$. The second integral is convergent by the argument you indicated.
Comment: How to choose the function $g(x)=x^{-1/2}$ to compare with $$f(x)=\frac{1}{\sqrt{x^{3}+x}}=x^{-1/2}\frac{1}{\sqrt{1+x^{2}}}\ ?$$ The binomial series $\left( 1+x\right) ^{\alpha }=\sum_{k=0}^{\infty }\binom{\alpha }{k}x^{k}$ yields for $\alpha =-1/2$
$$\frac{1}{\sqrt{1+x}}=1-\frac{1}{2}x+\frac{3}{8}x^{2}+\ldots .$$
Substituting $x^{2}$ for $x$, we get
$$\frac{1}{\sqrt{1+x^{2}}}=1-\frac{1}{2}x^{2}+\frac{3}{8}x^{4}+\ldots $$
Consequently,
$$f(x)=\frac{1}{\sqrt{x^{3}+x}}=x^{-1/2}-\frac{1}{2}x^{3/2}+\frac{3}{8}
x^{7/2}+\ldots .$$
The function $g(x)=x^{-1/2}$ is the first term of this expansion.
Best Answer
Since $x\log(x)$ is monotonic on $[1,\infty)$, let $f(x)$ be its inverse. That is, for $x\in[0,\infty)$ $$ f(x)\log(f(x))=x\tag{1} $$ Differentiating implicitly, we get $$ f'(x)=\frac{1}{\log(f(x))+1}\tag{2} $$ Then $$ \begin{align} \int_1^\infty\sin(x\log(x))\;\mathrm{d}x &=\int_0^\infty\sin(x)\;\mathrm{d}f(x)\\ &=\int_0^\infty\frac{\sin(x)}{\log(f(x))+1}\mathrm{d}x\tag{3} \end{align} $$ Since $\left|\int_0^M\sin(x)\;\mathrm{d}x\right|\le2$ and $\frac{1}{\log(f(x))+1}$ monotonically decreases to $0$, Dirichlet's test (Theorem 17.5) says that $(3)$ converges.