∫(2-x²)dx
=[2x-(1/3)x³]|
=(2√2-(1/3)*2√2]-(0-0)
=2√2-(2/3)√2
=(4/3)√2
∫1/[√(1-x)-1]dx
令√(1-x)=t ==> 1-x=t² ==> x=1-t² ==> dx=-2tdt
且,x=3/4时,t=1/2；x=1时,t=0
原式=∫[-2t/(t-1)]dt
=2∫[t/(t-1)]dt
=2∫[(t-1)+1]/(t-1)dt
=2∫[1+1/(t-1)]dt
=2t|+2∫[1/(t-1)]dt
=2[(1/2)-0]+2ln|t-1||
=1+2[ln(1/2)-0]
=1-2ln2
∫1/[e^x+e^(-x)]dx
=∫e^x/[(e^x)²+1]dx
=∫[1/(e^x)²+1]d(e^x)
=arctan(e^x)|
=limarctana-arctan1
=(π/2)-(π/4)
=π/4