We can play around with the substitution choice to ensure that the integral is expressed purely in terms of $t$.
From the substitution choice we infer:
$$t=\sqrt{x}+1$$
$$\implies t-1=\sqrt{x}$$
$$\implies (t-1)^2=x.$$
As GEdgar said, after using the substitution $t=\sqrt{x}+1$, you should obtain: $$dt=\frac{1}{2\sqrt{x}}dx$$
$$\implies 2\sqrt{x}dt=2(t-1)=dx.$$
We now make use of these equalities back in the integral:
$$\int \frac{x}{\sqrt{x}+1}dx$$
$$=\int \frac{2(t-1)^2(t-1)}{t}dt$$
$$=2\int \frac{(t-1)^3}{t}dt.$$
We can expand the numerator and then divide each term by $t$, like so:
$$2\int\frac{t^3-3t^2+3t-1}{t}dt$$
$$=2\int t^2-3t+3-\frac{1}{t}dt$$
Can you proceed from here?