An alternative way to calculate $\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx$ Consider the integral
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx
$$
By noting that
$$
\frac{x}{x^2+(1-x)^2}=
\frac{2x-1}{(2x-1)^2+1}+\frac{1}{(2x-1)^2+1}
$$
we deduce
$$
\int\frac{x}{x^2+(1-x)^2}\,dx=\frac{\ln(x^2+(1-x)^2)}{4}+\frac{1}{2}\arctan(2x-1)+C,
$$
so
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx=\frac{\pi}{4}.
$$
Is there an alternative way to calculate this integral?

A trick: use $\int_0^1f(x)dx=\int_0^1f(1-x)dx$ to average two versions of your integral, reducing it to$$\tfrac12\left(\int_0^1\tfrac{xdx}{x^2+(1-x)^2}+\int_0^1\tfrac{(1-x)dx}{x^2+(1-x)^2}\right)=\tfrac12\int_0^1\tfrac{dx}{x^2+(1-x)^2}.$$
