Integral of $\tan(\ln(x+1))/x$ How can I solve this integral?

$$\int{\frac{\tan(\ln(x+1))}{x}dx } $$

I tried substitution: $x+1 = e^t $, 

$$\int{ \frac{e^t \cdot \tan(t)}{e^t -1 }dt} $$ 

and U-substitution: $u = \tan(t), \ v=\ln(e^t-1) $, but it doesn't work. 

Thank you!
$|x|&lt;\frac{\pi}{2}$
$$\int\frac{e^t \tan t}{e^t-1}dt=\frac{1}{2}\int\tan t dt+\int(\frac{t}{2}+\frac{t}{e^t-1})\frac{ \tan t}{t}dt=$$
$$=\frac{1}{2}\ln|\cos t|+\int(\sum\limits_{k=0}^\infty \frac{B_{2k}}{(2k)!}t^{2k} )( \sum\limits_{k=0}^\infty (-1)^k\frac{2^{2k+2}(2^{2k+2}-1)B_{2k+2}}{(2k+2)!}t^{2k})dt$$
$$=\frac{1}{2}\ln|\cos t|+\sum\limits_{k=0}^\infty \frac{t^{2k+1}}{2k+1}\sum\limits_{v=0}^k (-1)^v\frac{2^{2v+2}(2^{2v+2}-1)B_{2v+2}B_{2k-2v}}{(2v+2)!(2k-2v)!}+C$$
Hint: Use a substitution $u=\ln(x+1)$ so that $du=\frac{1}{x}\,dx$. Can you solve the rest of it from here?

Edit: Follow the other dude :)