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$$\int_0^1 \frac{dx}{x^2+2x+5}\; $$

Is there a way to solve this integral without performing u-substitution and using the common integral for arctan?

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    Why two dx's? Anything special? – 2017-02-18
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    what about completing square and then using arctan? Avoiding the use of arctan could be a bit tough – 2017-02-18
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    Let me clarify that last comment further: are we allowed to use $\arctan$ at all, or do you want a completely different method? – 2017-02-18
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    Common arctan integral should be avoided. – 2017-02-18
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    To avoid $\arctan$, you could perform partial fraction decomposition. Of course, that will lead to complex $\ln$'s, which is really only a fancy rewriting of $\arctan$. – 2017-02-18

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Hint:

$$\frac1{x^2+2x+5}=\frac1{8i}\left(\frac1{x+1-4i}-\frac1{x+1+4i}\right)$$

Now use complex logarithms and the complex definition of arctan.

Alternatively, we can use the geometric series:

$$\frac1{x^2+2x+5}=\frac{1/4}{1+\frac{(x+1)^2}4}=\sum_{n=0}^\infty(-1)^n\frac{(x+1)^{2n}}{4^{n+1}}$$

And now it's a simple integration term by term, though this does not lend you closed forms nicely.

Remark: You should probably use arctangents when you can, because things get messy otherwise.

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    I just posted a comment about this myself, but it seems that the OP wants to avoid arctan – 2017-02-18
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    @BrevanEllefsen Fixed it? :-( – 2017-02-18
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    I mean, I guess so... what you've done is really more of a technicality XD but we will see if the OP likes it more. My guess is that if the OP isn't comfortable with arctan substitution he/she is DEFINITELY not going to be comfortable with the complex form, but only time will tell. I'll throw a +1 your way regardless :) – 2017-02-18
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    @BrevanEllefsen Lol, I definitely feel the same way. Complex arctangents coming from logarithms of complex numbers... not something I would enjoy. – 2017-02-18
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    (+1) for the first way forward. Note that the series expansion should be an alternating series. – 2017-02-18
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    @Dr.MVthanks for the catch. – 2017-02-18
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While it's trivial to transform this series into something much simpler, as we can see here
$$\int_0^1 \frac{dx}{x^2+2x+5}$$ $$=\int_0^1 \frac{dx}{(x+1)^2+4}$$ $$=\int_1^2 \frac{dx}{x^2+4} \operatorname*{=}^{x \to 1/x} \int_{1/2}^{1} \frac{dx}{4x^2+1}$$ The important thing is that the answer to your integral is $\frac{1}{2}\arctan(2)-\frac{\pi}{8}$ Note that there is NO simpler closed form for $\arctan(2)$

(except perhaps in terms of complex logarithms which is more of a technicality)