how this simplification was done in the integral? I'm stuck at simplification
Here is the Integral ,
$$I=\int\frac{x^2+x+1}{\sqrt{x^2+2x+3}} \ dx$$
To solve this ,
First let's Substitute, we got
$$x^2+2x+3=t^2$$
$$\implies {x}=\sqrt{(t^2-2)}-1$$
$$\implies dx=\frac{t}{\sqrt{t^2-2}}dt$$
Putting this back into the Integral followed by some simplification we have,
$$I=\int\sqrt{t^{2}-2}+\frac{1}{\sqrt{t^{2}-2}}-1 dt$$
Here, how they did it or how they "put back integral " and simplified it ? please, can anybody explain it step by step ?

Substitution is not the way to go here,
Write the integral as
$$=\int\sqrt{x^2+2x+3}dx-\int\frac{x+1}{\sqrt{x^2+2x+3}}dx-\int\frac{1}{\sqrt{x^2+2x+3}}dx$$
We have three types of integrals. The middle is a straightforward substitution.
The first can be done by parts,
$$\int\sqrt{x^2+2x+3}dx=x\sqrt{x^2+2x+3}-\int\frac{x^2+x}{\sqrt{x^2+2x+3}}dx$$
which gives,
$$2\int\sqrt{x^2+2x+3}dx=x\sqrt{x^2+2x+3}+\int\frac{x+3}{\sqrt{x^2+2x+3}}dx$$
and the final type,
$$\int\frac{1}{\sqrt{x^2+2x+3}}dx=\int\frac{1}{\sqrt{(x+1)^2+2}}dx$$ use a $$x+1=\sqrt{2}\sinh y$$ substitution.
