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I first completed the square:

$$int_0^1 frac1 {2+(x+1)^2},mathrm dx$$

Made the substitution $x+1=sqrt2 tan u$. Thus $dx=sqrt2sec^2udu$ substituting this in and changing the limits (please check that I have done this bit right).
$$frac{sqrt2}{2}int_{tan^{-1}frac1{sqrt2}}^{tan^{-1}frac2{sqrt2}},mathrm du$$
after using the identity $tan^2u+1=sec^2u$ and factoring out the constants.
Clearly this leads to a horrible result with many decimal places. Since this is meant to be done without a calculator, I suspect that I have done something wrong. Please help.

I don't understand what do you mean by horrible result

By proceeding with your solution or by evaluating indefinite integral and then applying limits we get

$$int frac1 {2+(x+1)^2}dx=frac{1}{sqrt{2}}arctanleft(frac{x+1}{sqrt{2}}right)$$

$$int_0^1 frac1 {2+(x+1)^2}dx=frac{1}{sqrt{2}}left[arctanleft(sqrt2 right)-arctanleft(frac{1}{sqrt{2}}right)right]$$

You can simplify this to $$frac{1}{sqrt{2}}operatorname {arccot}left(2 sqrt{2}right)$$

But it will still be a horrible result

we set $x+1=t$ and we get $dx=dt$ and our integral will be $frac{1}{2}intfrac{dt}{left(frac{t}{sqrt{2}}right)^2+1}$ and after $dt=sqrt{2}du$ we get
$$frac{1}{2}int frac{sqrt{2}du}{u^2+1}$$

Surprisingly we can take a very general approach here. Consider the function
$$I=I(x;a,b,c)=intfrac{mathrm dx}{ax^2+bx+c}$$ With the requirement that $4ac>b^2$. We may compute this by comleting the square:
$$I=intfrac{mathrm dx}{aleft(x+frac{b}{2a}right)^2+g}$$
Where $g=c-frac{b^2}{4a}$. Then preforming the substitution $x+frac{b}{2a}=sqrt{frac{g}{a}}tan u$,
$$I=sqrt{frac{g}{a}}intfrac{sec^2u,mathrm du}{gtan^2u+g}$$
Which simplifies to
$$I=frac{2u}{sqrt{4ac-b^2}}$$
$$I(x;a,b,c)=frac{2}{sqrt{4ac-b^2}}arctanfrac{2ax+b}{sqrt{4ac-b^2}},+C$$
So for your integral, we use $a=1,b=2,c=3$ to see that
$$int_0^1frac{mathrm dx}{x^2+2x+3}=I(1;1,2,3)-I(0;1,2,3)=frac1{sqrt2}operatorname{arccot}2sqrt2$$
Which is a very clean answer with infinite accuracy (as opposed to a decimal expansion)... What could be better?

Extra special Bit

If we define $$K_n=K(x;n;a,b,c)=intfrac{mathrm dx}{left[a(x+b)^2+cright]^{n+1}}$$
We can preform the substitution $w=x+b$ then integrate by parts with $mathrm dv=mathrm dw$ to see that $K_n$ satisfies the recurrence relation
$$K_n=frac{w}{2cn(aw^2+c)^n}+frac{2n-1}{2cn}K_{n-1}$$
With base case $K_0=I(w;a,0,c)$. Hence we have a solution to our recurrence:
$$K_n=frac1{4^nc^nsqrt{ac}}{2nchoose n}arctanleft[(x+b)sqrt{frac{a}{c}}right]+frac{x+b}{2c}sum_{k=0}^{n-1}frac{left[a(x+b)^2+cright]^{k-n}}{c^k(n-k)}prod_{i=1}^{k}frac{2n-2i+1}{2n-2i+2}$$
And if that in all of it's infinite precision is not beautiful, then I don't know what is.