求不定积分∫(1+x)/√(1-4x^2)dx 求不定积分∫1/(e^x+e^2x)dx

Answer 1

∫(1+x)/√(1-4x²) dx
= ∫dx/√(1-4x²) + ∫xdx/√(1-4x²)
= ∫dx/√[4(1/4-x²)] + (1/2)∫d(x²)/√(1-4x²)
= (1/2)∫dx/√(1/4-x²) + (1/2)(-1/4)∫d(1-4x²)/√(1-4x²)
= (1/2)arctan[x/(1/2)] - (1/8)*2√(1-4x²) + C
= (1/2)arctan(2x) - (1/4)√(1-4x²) + C
∫dx/(e^x+e^2x)
= ∫dx/[e^x*(1+e^x)]
= ∫(1+e^x-e^x)/[e^x*(1+e^x)] dx
= ∫1/e^x dx - ∫1/(1+e^x) dx
= ∫e^-x dx - ∫(1+e^x-e^x)/(1+e^x) dx
= -e^-x - ∫ dx + ∫e^x/(1+e^x) dx
= -e^-x - x + ∫d(1+e^x)/(1+e^x)
= -e^-x - x + ln|1+e^x| + C
= -e^-x + ln|1+e^x| - lne^x + C
= -e^-x + ln|(1+e^x)/e^x| + C
= -e^-x + ln|1+e^-x| + C