Evaluate $\int_0^1 \frac1 {x^2+2x+3}\,\mathrm dx$ I first completed the square:
$$\int_0^1 \frac1 {2+(x+1)^2}\,\mathrm dx$$
Made the substitution $x+1=\sqrt2 \tan u$. Thus $dx=\sqrt2\sec^2udu$ 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.
 A: 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
$$\int \frac1 {2+(x+1)^2}dx=\frac{1}{\sqrt{2}}\arctan\left(\frac{x+1}{\sqrt{2}}\right)$$
$$\int_0^1 \frac1 {2+(x+1)^2}dx=\frac{1}{\sqrt{2}}\left[\arctan\left(\sqrt2 \right)-\arctan\left(\frac{1}{\sqrt{2}}\right)\right]$$
You can simplify this to $$\frac{1}{\sqrt{2}}\operatorname {arccot}\left(2 \sqrt{2}\right)$$
But it will still be a horrible result
A: we set $x+1=t$ and we get $dx=dt$ and our integral will be $\frac{1}{2}\int\frac{dt}{\left(\frac{t}{\sqrt{2}}\right)^2+1}$ and after $dt=\sqrt{2}du$ we get
$$\frac{1}{2}\int \frac{\sqrt{2}du}{u^2+1}$$
A: Surprisingly we can take a very general approach here. Consider the function 
$$I=I(x;a,b,c)=\int\frac{\mathrm dx}{ax^2+bx+c}$$ With the requirement that $4ac>b^2$. We may compute this by comleting the square:
$$I=\int\frac{\mathrm dx}{a\left(x+\frac{b}{2a}\right)^2+g}$$
Where $g=c-\frac{b^2}{4a}$. Then preforming the substitution $x+\frac{b}{2a}=\sqrt{\frac{g}{a}}\tan u$, 
$$I=\sqrt{\frac{g}{a}}\int\frac{\sec^2u\,\mathrm du}{g\tan^2u+g}$$
Which simplifies to 
$$I=\frac{2u}{\sqrt{4ac-b^2}}$$
$$I(x;a,b,c)=\frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}\,+C$$
So for your integral, we use $a=1,b=2,c=3$ to see that 
$$\int_0^1\frac{\mathrm dx}{x^2+2x+3}=I(1;1,2,3)-I(0;1,2,3)=\frac1{\sqrt2}\operatorname{arccot}2\sqrt2$$
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 $$K_n=K(x;n;a,b,c)=\int\frac{\mathrm dx}{\left[a(x+b)^2+c\right]^{n+1}}$$
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:
$$K_n=\frac1{4^nc^n\sqrt{ac}}{2n\choose n}\arctan\left[(x+b)\sqrt{\frac{a}{c}}\right]+\frac{x+b}{2c}\sum_{k=0}^{n-1}\frac{\left[a(x+b)^2+c\right]^{k-n}}{c^k(n-k)}\prod_{i=1}^{k}\frac{2n-2i+1}{2n-2i+2}$$
And if that in all of it's infinite precision is not beautiful, then I don't know what is.
