integrate $\int_0^12\arctan(1+a\ln(x)) \ dx $ show $\int_0^12\arctan(1+a\ln(x)) \ dx = \frac{\pi}{2} - a -a^2 -a^3 + O(a^4) $ for $a \ll 1$
I've never seen this type of integral before, but it looks very familar to something to do with taylor, so I attempted this:
$I(x) = \displaystyle\int_0^12\arctan(1+a\ln(x)) \ dx$
$I(1) = I(1) + I'(1)(x-1) + I''(1)(x-1)^2/2! +...$ however this gets very very messy by the second derivative so I'm assuming there is something I'm missing. Any help, thanks.
 A: First, your integral is a function of $a$, not $x$: it gives a number for any $a$, choosing $x$ makes no sense.
$$I(a)=2\int_0^1 dx \arctan(1+a\ln x).$$
Second, your strategy to use the Taylor expansion was a good idea, but you should expand in $a$ instead of $x$ and around $0$ instead of $1$, because you need $a\ll 1$. So we seek for the coefficients in the expression
$$I(a)=I(0)+I'(0)a+\tfrac{1}{2!}I''(0)a^2+\tfrac{1}{3!}I'''(0)a^3+O(a^4).\quad(1)$$
We will first differentiate the integrand $f(a,x):=\arctan(1+a\ln x)$ w.r.t $a$:
$$\partial_a f(a,x)=\frac{\ln x}{1+(1+a\ln x)^2}$$
As you say it becomes messy, the trick is here to reexpress the new derivatives using the previous:
$$\partial_a^2f(a,x)=({\text{standard rules}})=\frac{-(\ln x)^2}{(1+(1+a\ln x)^2)^2}2(1+a\ln x)=-2(1+a\ln x)(\partial_a f)^2$$
Now this will save us a lot of work:
$$\partial_a^3f(a,x)=-2\ln x (\partial_a f)^2-2(1+a\ln x)\partial_a(\partial_a f)^2=-2\ln x(\partial_a f)^2-4(1+a\ln x)(\partial_a f)(\partial_a^2 f)$$
Next we just set $a=0$ and see what happens
$$f(0,x)=\arctan(1)=\tfrac{\pi}{4}$$
$$\partial_a f(0,x)=\tfrac{1}{2}\ln x$$
$$\partial_a^2 f(0,x)=-\tfrac{1}{2}(\ln x)^2$$
$$\partial_a^3 f(0,x)=\tfrac{1}{2}(\ln x)^3$$
Now we can compute the coefficients of the Taylor expansion
$$I(0)=2\int_0^1 dx f(0,x)=\tfrac{\pi}{2}$$
$$I'(0)=2\int_0^1 dx \partial_a f(0,x)=\int_0^1 dx \ln x =(x\ln x -x)_{\epsilon\to 0}^1=-1$$
$$I''(0)=2\int_0^1 dx \partial_a^2 f(0,x)=-\int_0^1 dx (\ln x)^2 =-(x(\ln x)^2 -2x\ln x+2x)_{\epsilon\to 0}^1=-2$$
$$I'''(0)=2\int_0^1 dx \partial_a^3 f(0,x)=\int_0^1 dx (\ln x)^3 =(x(\ln x)^3-3x(\ln x)^2+6x\ln x-6x)_{\epsilon\to 0}^1=-6$$
To keep it short, I did not explain how to do $\int dx (\ln x)^n$, please comment if you need further help on that!
Plugging everything into eq. (1) gives:
$$I(a)=\tfrac{\pi}{2}-a-a^2-a^3+O(a^4)$$
Edit: Thanks to @gammatester the prefactor in $I'''$ is fixed now :)
A: Expand $f(x,a) = 2 \arctan(1+a\ln x)$ into a series for small a
$$ g(x,a) = \frac{1}{2}\pi +a \ln x -\frac{1}{2}a^2\ln^2 x + \frac{1}{6} a^3 \ln^3 x - \frac{1}{20}a^5\ln^5x + O(a^6)$$
Then integrate termwise using the fact that (found by substitution $u = - \ln x$)
$$\int_0^1\ln^nx dx = (-1)^n\Gamma(n+1)$$
and you get 
$$I(a) \approx \int_0^1 g(x,a) dx = \frac{1}{2}\pi-a-a^2-a^3 + 6a^5$$
A: I have another approach to propose. Since "a" is small, directly expand the integrand as a Taylor series built around a=0. This gives  
Pi / 2 + a Log[x] - 1/2 a^2 Log[x]^2 + 1/6 a^3 Log[x]^3 - 1/20 a^5 Log[x]^5 + ...
Now, integrate with repect to "x". As shown by flonk, this then asks to compute integrals of Log[x]^n. I shall not cover this point here unless, as flonk already mentioned, you ask for it.  
As a result of the integration between 0 and 1, we the arrive to      
Pi / 2 -a - a^2 - a^3 + 6 a^5 + ...
