∵f(x)=ln(x+√(1+x²))∴f'(x)=[ln(x+√(1+x²))]'=(1+x/√(1+x²))/(x+√(1+x²))=((x+√(1+x²))/√(1+x²))/(x+√(1+x²))=1/√(1+x²)故∫xf'(x)dx=∫xdx/√(1+x²)=(1/2)...
求不定积分 ∫ xf'(x)dx, 其中f(x)=ln（x+根号1+x^2）
∵f(x)=ln(x+√(1+x²))∴f'(x)=[ln(x+√(1+x²))]'=(1+x/√(1+x²))/(x+√(1+x²))=((x+√(1+x²))/√(1+x²))/(x+√(1+x²))=1/√(1+x²)故∫xf'(x)dx=∫xdx/√(1+x²)=(1/2)...