Another way of proving :$\int_{0}^{1}{x-x^3+x^5-x^7\over (1+x^4)\ln{x}}dx=-\ln{2}$ Prove that
$$\int_{0}^{1}{x-x^3+x^5-x^7\over (1+x^4)\ln{x}}dx=-\ln{2}$$
My try
$x-x^2+x^3+x^5-x^7=x(1-x^2)+x^5(1-x^2)=x(1+x^4)(1-x^2)$
$$\int_{0}^{1}{x(1+x^4)(1-x^2)\over (1+x^4)\ln{x}}dx$$
Applying Frullani theorem 
$$\int_{0}^{1}{x-x^3\over \ln{x}}dx$$
$$\int_{0}^{1}{x-x^3\over \ln{x}}dx=-\ln{2}$$

Maybe some one will be interested in that
We can show
$$ \int^1_0 \frac{x^a-1 }{\log x} dx =\log \left ( a+1\right)$$
Using differentiation under the integral sign. 
Generally
$$ \int^1_0 \frac{x^a-x^b }{\log x} dx =\int^1_0 \frac{x^a-1-(x^b-1)}{\log x} dx =\log \left ( \frac {a+1}{b+1}\right)$$
Hence
$$ \int^1_0 \frac{x-x^3 }{\log x} dx =\log \left ( \frac {2}{4}\right)=-\log 2$$
