How to solve this trig integral? I tried to solve this integral but I don't know how. Can someone help me please?
$$\int _0^2\frac{\arctan \left(3x\right)}{x}dx\:$$
Thanks for any possible response.
Edit: My trial:
1. Substitution:
$$\int _0^2\frac{\arctan \left(3x\right)}{x}dx\: = 9\int _0^2\:\:\frac{\arctan \left(u\right)}{u}du = (2.IBP => dv = \frac{1}{u}, v = \log _e\left(u\right); q = \arctan \left(u\right), dq= \frac{1}{1+u^2}) = 9(\log _e\left(u\right)\arctan \left(u\right) -  \int _0^2 \frac{1}{1+u^2}\log _e\left(u\right)) = (3.IBP => dv = \frac{1}{1+u^2}) , v =  \arctan \left(u\right); q = \log _e\left(u\right), dq= \frac{1}{u}) = 9(\log _e\left(u\right)\arctan \left(u\right) -( \arctan \left(u\right)\log _e\left(u\right) - \int _0^2 \arctan \left(u\right)\frac{1}{u} $$
As you can see I arrived at the same thing.
 A: $\frac{\arctan x}{x}$ does not have an elementary antiderivative, and the radius of convergence of its Taylor series at $x=0$ is just one, so we need a couple of tricks. 


*

*$$\int_{0}^{2}\frac{\arctan(3x)}{x}\,dx = \int_{0}^{6}\frac{\arctan x}{x}\,dx = \log(6)\arctan(6)-\int_{0}^{6}\frac{\log(x)}{1+x^2}\,dx$$

*$$ \int_{0}^{6}\frac{\log(x)}{1+x^2}\,dx = -\int_{1/6}^{+\infty}\frac{\log(x)}{1+x^2}\,dx=\int_{0}^{1/6}\frac{\log(x)}{1+x^2}\,dx$$

*$$ \int_{0}^{1/6}x^{2k}\log(x)\,dx = -\frac{1+(2k+1)\log 6}{6^{2k+1}(2k+1)^2}$$
give:

$$ \int_{0}^{2}\frac{\arctan(3x)}{x}\,dx = \color{red}{\frac{\pi}{2}\log(6)+\sum_{k\geq 0}\frac{(-1)^k}{6^{2k+1}(2k+1)^2}}.$$

A: Enforce the substitution $u=3x$ to get:
$$\int_{0}^{6} \frac{\arctan u}{u} du$$
Now use the identity found in Wikipedia https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions (scroll down to logarithmic forms) : 
$$\arctan x=\frac{1}{2}i [\ln(1-ix)-\ln(1+ix)]$$
So
$$\frac{\arctan (x)}{x}= \frac{-1}{2} [\frac{\ln (1-ix)}{ix} - \frac{\ln (1+ix)}{ix}]$$
Use this to transform the integral to:
$$\frac{-1}{2} \int_{0}^{6} [\frac{\ln (1-ix)}{ix} - \frac{\ln (1+ix)}{ix}] dx$$
Now the substitution $t=ix$, $-idt=dx$ to get:
$$\frac{i}{2} \int_{0}^{6i} [\frac{\ln (1-t)}{t} -\frac{\ln (1+t)}{t}] dt$$
Let's look at:
$$\int_{0}^{6i} \frac{\ln (1+t)}{t} dt$$
Enforce the substitution $t=-w$:
$$\int_{0}^{-6i} \frac{\ln (1-w)}{w} dw$$
Note the dilogarithm is defined as:
$$\textrm{Li}_2(z)=\int_{z}^{0} \frac{\ln (1-t)}{t} dt$$
So this integral is equal to:
$$-\textrm{Li}_2(-6i)$$
The other integral is,
$$\int_{0}^{6i} \frac{\ln (1-t)}{t} dt=-\textrm{Li}_2(6i)$$
So your integral is equal to,
$$\frac{i}{2}[-\textrm{Li}_2(6i)+\textrm{Li}_2(-6i)]$$
Please correct me if you see anything wrong.
A: I would suggest integration by parts. Hard to integrate arctan(3x) when its derivative isn't popping up anywhere in the integrand. So integration by parts using arctan(3x) = u, dx/x = dv. I realize this will get an ln. Might take a couple of IBP before the solution falls out.
A: Rewrite the integral like this: 
$$\int _0^2 \frac {1}{x}\arctan(3x)\mathop{\mathrm{d}x}$$
Then we just use integration by parts to solve. Here,I'll get you started: 
$u = \arctan(3x)$            
$\mathop{\mathrm{d}u} = \dfrac{1} {1+x^2}\mathop{\mathrm{d}x}$
$\mathop{\mathrm{d}v} = \dfrac {1}{x}\mathop{\mathrm{d}x}$ 
$v  = \ln x$
Then:
$$\int u \mathop{\mathrm{d}v} = uv - \int v \mathop{\mathrm{d}u} = \arctan(3x) \ln x - \int_0^2\frac{\ln x}{1+x^2} \mathop{\mathrm{d}x}$$
We now repeat until we obtain the desired integral. 
