<s>[INST] Compute $\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$ $$\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$$
Then the question ask me to change it into 
$$\int_0^3 \frac{(x-3)^4}{x^4+(x-3)^4} \,dx$$
Then how to evaluate it 
if let $$u=x-3$$
$$du=dx$$
$$x^4 =(u+3)^4$$
$$\int_3^0 \frac{u^4}{(u+3)^4+u^4} \,du$$
it is still the same
so what should i do? 
 [/INST] Let,
$$I=\int_0^3 \frac{x^4}{x^4+(x-3)^4} \,dx$$
Let $x=3-u$ and change dummy variable back to $x$.
$$I=\int_{0}^{3} \frac{(3-x)^4}{x^4+(x-3)^4} dx$$
$$=\int_{0}^{3}  \frac{(x-3)^4}{x^4+(x-3)^4} dx$$
Add this to the first form of $I$.
$$2I=\int_{0}^{3} 1 dx$$
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