how this simplification was done in the integral?

Question

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 ?

Answer 1

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.

Answer 2

These are the key steps of the substitution:

$$ \frac{x^2+x+1}{\sqrt{x^2+2x+3}}dx=\frac{x^2+2x+3-x-2}{t}dx= $$ $$ =\frac{t^2-1-\sqrt{t^2-2}}{t}\frac{t}{\sqrt{t^2-2}}dt= $$ $$ =\left(\frac{t^2-2}{\sqrt{t^2-2}}+\frac{1}{\sqrt{t^2-2}}-1\right)dt = $$ $$ \left(\sqrt{t^{2}-2}+\frac{1}{\sqrt{t^{2}-2}}-1\right) dt $$

Answer 3

this simplification is not correct, let me do the simplification with this substitution.

$$t^{2}=x^{2}+2 x+3=x^{2}+2 x+1+2=(x+1)^{2}+2 \Rightarrow \sqrt{t^{2}-2}-1=x.$$

$$\frac{t}{\sqrt{t^{2}-2}} d t=d x.$$

$$t^{2}=x^{2}+2 x+3=x^{2}+x+1+x+2 \Rightarrow t^{2}-x-2=t^{2}-\sqrt{t^{2}-2}-1=x^{2}+x+1.$$

$$\displaystyle I=\int \frac{x^{2}+x+1}{\sqrt{x^{2}+2 x+3}} d x=\int \frac{\left(t^{2}-\sqrt{t^{2}-2}-1\right) t}{t \sqrt{t^{2}-2}} d t=\int \left(\frac{t^{2}}{\sqrt{t^{2}-2}}+\frac{1}{\sqrt{t^{2}-2}}-1 \right) d t. $$

But it is clear that.

$$\sqrt{t^{2}-2} \neq \frac{t^{2}}{\sqrt{t^{2}-2}}.$$

Answer 4

If you let $x=\sqrt{t^2-2}-1,$ then $\mathrm{d}x=\frac{t}{\sqrt{t^2-2}}\,\mathrm{d}t,$ $x^2+2x+3=t^2,$ and $x^2+x+1=x^2+2x+3-x-2=t^2-\left(\sqrt{t^2-2}-1\right)-2=t^2-1-\sqrt{t^2-2}.$ Therefore, $$\int\frac{x^2+x+1}{\sqrt{x^2+2x+3}}\,\mathrm{d}x=\int\frac{t^2-1-\sqrt{t^2-2}}{\sqrt{t^2}}\frac{t}{\sqrt{t^2-2}}\,\mathrm{d}t=\int\frac{t^2-1-\sqrt{t^2-2}}{\sqrt{t^2-2}}\sqrt{t^2}\,\mathrm{d}t$$ $$=\int\frac{t^2-1}{\sqrt{t^2-2}}\sqrt{t^2}\,\mathrm{d}t-\int\sqrt{t^2}\,\mathrm{d}t=\int\sqrt{t^2-2}\sqrt{t^2}\,\mathrm{d}t+\int\frac1{\sqrt{t^2-2}}\sqrt{t^2}\,\mathrm{d}t-\int\sqrt{t^2}\,\mathrm{d}t=\int\sqrt{t^2-2}|t|\,\mathrm{d}t+\int\frac1{\sqrt{t^2-2}}|t|\,\mathrm{d}t-\int|t|\,\mathrm{d}t.$$

This is the correct way to do the substitution.