Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx $ Problem : 
Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx$.
My approach : 
\begin{align}
&\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx \\
=& \ln(1^x +2^x +3^x +6^x) \frac{x^2}{2} - \int^1_{-1} \frac{1}{1^x+2^x+3^x+6^x}(2^x\log2+3^x \log3+6^x \log 6 )\frac{x^2}{2} \,dx
\end{align}
[By using by parts]
Is it the correct method of solving this, please suggest, will be of great help, thanks. 

$$1+2^x+3^x+6^x=(1+2^x)(1+3^x)$$
Now $\ln(AB)=\ln A+\ln B$
Apply $\displaystyle\int_p^qf(x)\ dx=\int_p^qf(p+q-x)\ dx$
in $I=\displaystyle\int_{-1}^1x\ln(1+a^x)\ dx=\int_{-1}^1(-x)\ln(1+a^{-x})\ dx=\int_{-1}^1(-x)\{\ln(1+a^x)-x\ln a\}dx$
$\implies\displaystyle I=-I+\ln a\int_{-1}^1x^2\ dx$
