Calculate $\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln\left(x+1-\frac{1}{x}\right) \mathrm{dx}$ $$\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln\left(x+1-\frac{1}{x}\right) \mathrm{dx}$$
Insane integral!  So far I have tried to complete the square for the denominator then substitute and use taylor series for the natural logarithm about x=0.  Is this integral possible?!
 A: Here's a hint you may find helpful:
$$\int_{1}^{\phi}\dfrac{x^2+1}{x^4-x^2+1}ln\Big(x-\dfrac{1}{x}+1\Big)dx=\int_{1}^{\phi}\dfrac{1+\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}-1}ln\Big(x-\dfrac{1}{x}+1\Big)dx$$
Substituting $x-\dfrac{1}{x}=t$ gives:
$$x-\dfrac{1}{x}=t\Rightarrow\Big(1+\dfrac{1}{x^2}\Big)dx=dt$$
Therefore the integral becomes:
$$\int_{0}^{\phi-\frac{1}{\phi}}\dfrac{ln(t+1)}{t^2+1}dt$$
I hope you can proceed from here.
A: I assume that $\phi$ is the golden ratio.  Consider $u=x-\frac{1}{x}$ so that $\frac{x^2}{x^2+1} du=dx$:
$$I=\int_0^1 \frac{x^2}{x^2+1} \cdot \frac{x^2+1}{x^4-x^2+1} \ln{\left(1+u\right)} \; du$$
$$I=\int_0^1 \frac{x^2}{x^4-x^2+1} \ln{\left(1+u\right)} \; du$$
$$I=\int_0^1 \frac{1}{x^2-1+\frac{1}{x^2}} \ln{\left(1+u\right)} \; du$$
$$I=\int_0^1 \frac{1}{u^2+1} \ln{\left(1+u\right)} \; du$$
Now, let $u=\tan{t}$:
$$I=\int_0^{\frac{\pi}{4}}  \ln{\left(1+\tan{t}\right)} \; dt$$
Then, substitute $w=\frac{\pi}{4}-t$
$$I=\int_0^{\frac{\pi}{4}}  \ln{\left(1+\tan{\left(\frac{\pi}{4}-w\right)}\right)} \; dw$$
Use the tangent angle addition formula:
$$I=\int_0^{\frac{\pi}{4}}  \ln{\left(1+\frac{1-\tan{w}}{1+\tan{w}}\right)} \; dw$$
$$I=\int_0^{\frac{\pi}{4}}  \ln{2} \; dw-\int_0^{\frac{\pi}{4}} \ln{\left(1+\tan{w}\right)} \; dw$$
Remember that the second integral is $I$:
$$2I=\int_0^{\frac{\pi}{4}}  \ln{2} \; dw$$
$$I= \frac{\pi \ln{2}}{8}$$
