Question

1.∫x/√（a^2+x^2)dx

Answer 1

1、∫ x/√(a²+x²) dx

= ∫ d(x²/2)/√(a²+x²)

= (1/2)∫ d(x²+a²)/√(a²+x²)

= (1/2) * 2√(a²+x²) + C

= √(a²+x²) + C

2、∫ 1/√[x(a²-x²)] dx 无解

是∫ 1/[x√(a²-x²)] dx?

令x = a*sinz 则 dx = a*cosz dz

cscz = a/x,cotz = √(a²-x²)/x

∫ 1/[x√(a²-x²)] dx = ∫ a*cosz/(a*sinz*a*cosz) dz

= (1/a)∫ cscz dz

= (1/a) * -ln|cscz + cotz| + C

= (-1/a) * ln|[a+√(a²-x²)] / x| + C

= (-1/a) * [ln|a+√(a²-x²)| - ln|x|] + C

= [ln|x| - ln|a+√(a²-x²)|] / a + C

3、∫ √(e^x-2) dx

令u = √(e^x-2),e^x = 2+u²,x = ln(2+u²),dx = 2u/(2+u²) du

原式 = 2∫ u²/(2+u²) du

= 2∫ (2+u²-2)/(2+u²) du

= 2∫ du - 4∫ du/(2+u²)

= u - 4*(1/√2)*arctan(u/√2) + C

= √(e^x-2) - 2√2*arctan[√(e^x-2) / √2] + C