Question

求定积分∫下限0上限2 x／（x∧2-2x＋2）∧2dx

Answer 1

问：求定积分∫下限0上限2 x／（x∧2-2x＋2）∧2dx

答：您好，很高兴为您键好孝解答∫xdx/(x^2-2x+2)^2=∫袜改(x-1)dx/[(x-1)^2+1]^2 +∫稿稿d(x-1)/[(x-1)^2+1]^2(x-1)=tanu=∫tanudu+∫du= -ln|cosu|+u+C=ln|√(x^2-2x+2)|+arctan(x-1)+C∫[0,2]xdx/[(x^2-2x+2)=ln2+π/4