<s>[INST] 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. 
 [/INST] $$I=\int_{-1}^0f(x)dx+\int_0^1f(x)dx$$
$$=\int_0^1(f(-x)+f(x))dx$$
$$=\int_0^1(-f(x)+x\ln(6^x)+f(x))dx$$
$$=\ln(6)\int_0^1x^2dx$$
$$=\frac{\ln(6)}{3}.$$
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