作业帮∫x[ln(x²+2)-ln(2x+1)]dx
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/14 16:10:14
![∫x[ln(x²+2)-ln(2x+1)]dx](/uploads/image/z/4055914-10-4.jpg?t=%E2%88%ABx%5Bln%28x%26%23178%3B%2B2%29-ln%282x%2B1%29%5Ddx)
∫x[ln(x²+2)-ln(2x+1)]dx
∫x[ln(x²+2)-ln(2x+1)]dx
∫x[ln(x²+2)-ln(2x+1)]dx
∫x[ln(x²+2)-ln(2x+1)]dx
=∫xln(x²+2)dx-∫xln(2x+1)dx
=(1/2)∫ ln(x²+2)d(x²)-(1/2)∫ ln(2x+1)d(x²)
=(1/2)x²ln(x²+2)-(1/2)∫ x²/(x²+2)d(x²)-(1/2)x²ln(2x+1)+(1/2)∫ 2x²/(2x+1)dx
=(1/2)x²ln(x²+2)-(1/2)∫ (x²+2-2)/(x²+2)d(x²)-(1/2)x²ln(2x+1)+(1/4)∫ (4x²-1+1)/(2x+1)dx
=(1/2)x²ln(x²+2)-(1/2)∫1dx+∫ 1/(x²+2)d(x²)-(1/2)x²ln(2x+1)+(1/4)∫(2x-1)dx+(1/4)∫1/(2x+1)dx
=(1/2)x²ln(x²+2)-x/2+ln(x²+2)-(1/2)x²ln(2x+1)+(1/4)(x²-x)+(1/8)ln(2x+1)+C
=(1/2)x²ln(x²+2)+ln(x²+2)-(1/2)x²ln(2x+1)+(1/4)x²-(3/4)x+(1/8)ln(2x+1)+C