①は部分積分
∫tan^-1xdx=∫(x')tan^(-1)x dx
=x tan^(-1)x-∫x{1/(1+\(x^{2}\))}dx
=x tan^(-1)x-(\(\frac{1}{2}\))∫{2x/(1+\(x^{2}\))}dx
=x tan^(-1)x-(\(\frac{1}{2}\))log|1+\(x^{2}\)|
=x tan^(-1)x-log\(\sqrt{\quad}\)(1+\(x^{2}\))

②は、∫1/(\(x^{2}\)+1)dx=arctan(x)を変数変換して求めればよい。
∫1/{(x-p\()^{2}\)+\(q^{2}\)} dx=(\(\frac{1}{q}\))arctan{(x-p)/q}を参考に！
被積分関数は、3/(\(x^{2}\)-x+1)=3/{(x-\(\frac{1}{2}\)\()^{2}\)+(\(\sqrt{\quad}\)\(\frac{3}{2}\)\()^{2}\)}と変形。