Question

求不定积分∫xe^2x*dx 求定积分∫（1,0）dx/2+√x

Answer 1

∫xe^2xdx
=1/2∫xe^2xd2x
=1/2∫xde^2x
=(1/2)xe^2x-1/2∫e^2xdx
=(1/2)xe^2x-1/4∫e^2xd2x
=(1/2)xe^2x-(1/4)e^2x+C
∫(1,0)dx/2+√x
令√x=a
x=a²
dx=2ada
x=1,a=1
x=0,a=0
原式=∫(1,0)ada/(2+a)
=∫(1,0)(2+a-2)da/(2+a)
=∫(1,0)[1-2/(2+a)]da
=∫(1,0)[1-2/(2+a)]d(2+a)
=(2+a)-2ln(2+a)(1,0)
=(3-2ln3)-(2-2ln2)
=1-2ln3+2ln2