それぞれ置換で攻めます。

1.
x=tanθ と置くと、dx=dθ/(cosθ\()^{2}\)

∫[0→1] xarctanx dx
= ∫[0→π/4] θtanθ/(cosθ\()^{2}\) dθ

ここで、不定積分
∫θtanθ/(cosθ\()^{2}\) dθ
= ∫θsinθ/(cosθ\()^{3}\) dθ
= θ/( 2(cosθ\()^{2}\) ) - ∫dθ/( 2(cosθ\()^{2}\) )
　※∫-f'(t)/( f(t\()^{3}\) )dt = 1/( 2f(t\()^{2}\) ) + C を利用した部分積分
= θ/( 2(cosθ\()^{2}\) ) - \(\frac{1}{2}\)・tanθ + C

定積分の計算に戻って
∫[0→π/4] θtanθ/(cosθ\()^{2}\) dθ
= [ θ/( 2(cosθ\()^{2}\) ) - \(\frac{1}{2}\)・tanθ ][0→π/4]
= π/4 - \(\frac{1}{2}\) …(答え)


2.
x=\(e^{t}\) と置くと、dx=\(e^{t}\)・dt

∫[e\(\vec{e}\\()^{2}\)] 1/(xlogx) dx
= ∫[1→2] 1/(t\(e^{t}\))・\(e^{t}\) dt
= ∫[1→2] d\(\frac{t}{t}\)
= [ logt ][1→2]
= log2 …(答え)