![∫x[ln(x²+2)-ln(2x+1)]dx,第1張](data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw== "∫x[ln(x²+2)-ln(2x+1)]dx,第1張")
∫x[ln(x² 2)-ln(2x 1)]dx網友廻答：
- ∫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
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![∫x[ln(x²+2)-ln(2x+1)]dx,第2張](/statics/default/images/niming.jpg "∫x[ln(x²+2)-ln(2x+1)]dx,第2張")
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