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Let $f_n:]0, infty[to mathbb R$ and $f_n(x):=fracnsqrtx sin(x)1+nx^2$

Show that $f_n$ converges in $L^p$ to $f$ where $f(x):=x^-frac32sin(x)$ and $p in [1,2[$

My idea:

beginalign
lVert f_n-frVert_p^p&=int_0^inftyleftlvertfracnsqrtx sin(x)1+nx^2-x^-frac32sin(x)rightrvert^pdxleqint_0^inftyleftlvertfracnsqrtx1+nx^2-x^-frac32rightrvert^pdx \
&=int_0^inftyleftlvertfracnsqrtx-x^-frac32(1+nx^2)1+nx^2rightrvert^pdx=int_0^inftyleftlvertfracnsqrtx-x^-frac32+nx^-frac121+nx^2rightrvert^pdx
endalign

Am I on the right track? How do I continue from here?

First notice that

$$fracnsqrtxsin xn+x^2 = fracsqrtxsin xfrac1n+x^2 xrightarrowntoinfty x^-3/2sin x$$
pointwise.

Then notice that

$$left|fracsqrtxsin xfrac1n+x^2right|^p le frac1x^p/2chi_(0,1](x) + frac1x^3p/2chi_[1,infty)(x) =: g(x) in L^1(0,infty)$$
and also $|x^-3/2sin x|^p$ is dominated by the same integrable function $g$.

Hence

$$left|fracsqrtxsin xfrac1n+x^2 - x^-3/2sin xright|^ple 2^p g(x)$$
so Lebesgue dominated convergence theorem yields
$$lim_ntoinftyint_0^infty left|fracsqrtxsin xfrac1n+x^2 - x^-3/2sin xright|^p,dx = int_0^inftylim_ntoinfty left|fracsqrtxsin xfrac1n+x^2 - x^-3/2sin xright|^p,dx = int_0^infty 0,dx = 0$$

Let $f(x)= x^-3/2 sin, x$. Then $f_n(x) to f(x)$ for all $x$. $frac t 1+tx^2$ is an increasing function of $t$ so $frac n 1+nx^2 sqrt x$ is an increasing sequence of non-negative measurable functions converging at every point to $x^-3/2$. By Monotone Convergence Theorem see that $int_1^infty |f_n(x)-f(x)|dx to 0$ (because $|sin,x | leq 1$). This proves that the integral from $1$ to $infty$ tends to $0$ when $p=1$. For the general case not the simple

Lemma
If $0leq g_n$ and $g_n$ increases to $g$ with $int g^p <infty$ then $int |g_n-g|^p to 0$.

This lemma follows from Dominated Convergence Theorem

For $x <1$ note that $frac n 1+nx^2 sqrt x leq frac 1 2$, so Dominated Convergence Theorem shows that $int_0^1 |f_n(x)-f(x)|^pdx to 0$.

There is a small mistake in your calculation:
$$ n x^1/2 - x^-3/2 (1+nx^2) = n x^1/2 - x^-3/2 colorred- n x^colorred+ 1/2 = - x^-3/2 .$$
Therefore, the integral you actually end up with is
$$ int limits_0^infty fracmathrmdxx^3p/2 (1+nx^2)^p , $$
which is divergent at the lower limit. This is because the estimate $lvertsin(x)rvertleq 1$ is not good enough. Instead we should use $lvertsin(x)rvertleq x$, which is accurate at the lower limit. It is of course terribly bad for large $x$, but that is irrelevant since convergence at infinity is ensured by the additional $x^-2p$ term anyway.

With these changes we find
$$ lVert f_n - f rVert_p^p leq int limits_0^infty fracmathrmdxx^p/2 (1+nx^2)^p stackrelsqrtn x = t= n^-frac2-p4 int limits_0^infty fracmathrmdtt^p/2 (1+t^2)^p, .$$
While the remaining integral can be computed explicitly, it is of course sufficient to check that it converges.