Question

求不定积分∫xarcsinx/√（1-x^2)dx

Answer 1

简单计算一下即可，答案如图所示

Answer 2

∫x*arcsinx/√(1-x²)dx =∫arcsinx*[x/√(1-x²)dx] =∫arcsinxd[-√(1-x²)] =-√(1-x²)arcsinx+∫√(1-x²)d(arcsinx) =-√(1-x²)arcsinx+∫√(1-x²)*1/√(1-x²)dx =-√(1-x²)arcsinx+x+C