Question

∫(2x³+根号x-x/2)的不定积分

Answer 1

问：∫(2x³+根号x-x/2)的不定积分

答：∫(2x³+根号x-x/2)的不定积分是∫x/√(x-x²) dx= ∫x/√[(1/4)-(x-1/2)²] dx，令x-1/2=(1/2)sinθ，dx=(1/2)cosθdθ= ∫(1/2+1/2*sinθ)/[(1/2)cosθ] * (1/2)cosθdθ= (1/2)∫(1+sinθ)dθ= (1/2)(θ-cosθ) + C，由于sinθ=2x-1，cos²θ=1-(2x-1)²=1-(4x²-4x+1)=4x-4x²=>cosθ=2√(x-x²)= (1/2)arcsin(2x-1) - √(x-x²) + C

问：∫(2cosx+x²+(1+x²)╱1)的不定积分

答：∫ [2x/(x^2+x+1) ]dx= ∫ [(2x+1)/(x^2+x+1) ]dx -∫ dx/(x^2+x+1)=ln|x^2+x+1| -∫ dx/(x^2+x+1)considerx^2+x+1 = (x+1/2)^2 + 3/4letx+1/2 = (√3/2)tanydx = (√3/2)(secy)^2 dy∫dx/(x^2+x+1)=(2√3/3)∫ dy=(2√3/3)y + C'=(2√3/3)arctan[(2x+1)/√3] + C'∫ [2x/(x^2+x+1) ]dx=ln|x^2+x+1| -∫ dx/(x^2+x+1)=ln|x^2+x+1| -(2√3/3)arctan[(2x+1)/√3] + C